73
$\begingroup$

She visits third class and is $8$ years old (you can imagine how ashamed I felt when I said so to her). I helped her with lots of maths stuff today already but this is very unknowable for me. Sorry it's in German but I have translated it :)

enter image description here

It's saying "Each letter represents a digit. Determine them". First question, what is "them"? The letters I guess?

How shall I determine them when they are unknown? Or is it simply $A=1, B=2, C=3, D=4, E=5, F=6, G=7, H=8, K=11, L=12, M=13, N=14$?

Alright...

With this we gave a) the first try:

enter image description here

It doesn't seem to make sense to set $A=1, B=2, ...$

Or we did something wrong.. Any ideas how this could be solved? :s

$\endgroup$
  • 8
    $\begingroup$ Wikipedia calls this kind of puzzle "Verbal arithmetic". See en.wikipedia.org/wiki/Verbal_arithmetic $\endgroup$ – awkward Mar 4 '17 at 21:16
  • 22
    $\begingroup$ @awkward: the better term is Cryptarithm. "Verbal arithmetic" is way too vague, like could wrongly be inferred to apply to "If it takes seven men six days to dig three trenches..." $\endgroup$ – smci Mar 5 '17 at 2:26
  • 3
    $\begingroup$ The entirety of the directions says, "Each letter represents a digit. Determine it.", which would seem to answer your first question. $\endgroup$ – Daniel R. Collins Mar 5 '17 at 2:56
  • 3
    $\begingroup$ You ask It's saying "Determine them". First question, what is "them"? The letters I guess? Did you not read “jeder Buchstabe steht fuer eine Ziffer”? (every letter stands for a digit) $\endgroup$ – Mawg Mar 6 '17 at 9:31
  • 7
    $\begingroup$ It says "jeder Buchstabe steht fuer eine Ziffer". Ziffer is digit. K can't be 11, that's two digits. $\endgroup$ – RemcoGerlich Mar 6 '17 at 10:15

15 Answers 15

40
$\begingroup$

1.This is a homework for a third grade girl, so we need to think as a kid but not as an advanced expert, in this case we will be able to explain to this kid step by step how we solve the problem and he will understand the solution in an easy way.

2.I do not think that the purpose of the teacher of a third grade was to make all letters equal to 0 because it will not make any sense to the children and they will not learn anything from this case, so I think that this is not a valid solution at all.

3.FYI: the solutions must be from right to left because this is how children learn to do calculations.

4.The three operations seems to be independents because they are no common letters between them.

5.I have started to solve the third operation (c):

image of the third operation (c)

The trick is to firstly we give the value 8 to C then we can find B because 8+8+8=24 which means we write 4 and we put 2 on the top of the second column. Therefore B = 4 then we replace B by 4 in the second column and in the result. In the second column we get 2+4+4+4=14 we write 4 and we put 1 on the top of the last column, in the last column we have the result 4 and we already have 1 on the top so A will be 1 (1+1+1+1=4)

SO A=1 ,B=4 ,C=8

6.For the second operation (b): image of the second operation (b)

We give the value 1 to M then L= M+M = 1+1 = 2 so K = L+L = 2+2 = 4 in the end N = K+K = 4+4 = 8 therefore M=1, L=2, K=4, N=8 .Please notice if we give to M a value of 2,4,6,7,9 so the result will be on more then 3 digits (e.g. M=2 the result will be 1684) Therefore the solutions for M are 1,3,5,8.

$\endgroup$
  • 3
    $\begingroup$ I think you're right. It's a pretty flawed question but the only thing that makes sense is to treat them independently. $\endgroup$ – Erick Wong Mar 4 '17 at 23:28
  • 16
    $\begingroup$ @ErickWong there are twelve letters used. Some digit needs to be used multiple times. $\endgroup$ – quid Mar 5 '17 at 0:32
  • 2
    $\begingroup$ Did you even try other things for $M$ besides 1 and 2?? $\endgroup$ – user21820 Mar 5 '17 at 10:30
  • 3
    $\begingroup$ @user21820 No I did not, You are right, there is more than one solution, there are four solutions: M=1, M=3, M=5, M=8. Good catch. $\endgroup$ – Billel Hacaine Mar 5 '17 at 10:41
  • 2
    $\begingroup$ @BillelHacaine: That's right; see my answer where I said there are four solutions! $\endgroup$ – user21820 Mar 5 '17 at 10:41
38
$\begingroup$

Each letter stands for a digit, but not all of them are different. For example, if $A=1, B=4$ and $C=8$, that makes the last equation correct. It's more than standard third grade math, it's a puzzle to try to figure out which letters are which digits.

$\endgroup$
  • 36
    $\begingroup$ That's one answer. I'd prefer to just say all letters are zero. Much faster that way, and all the equations work out just as well. $\endgroup$ – penchant Mar 7 '17 at 1:24
  • 1
    $\begingroup$ @tr3buchet: Usually when you mean zero, you write $0$, not $000$. $\endgroup$ – tomasz Mar 7 '17 at 9:57
  • $\begingroup$ @tr3buchet From my answer:"I do not think that the purpose of the teacher of a third grade was to make all letters equal to 0 because it will not make any sense to the children and they will not learn anything from this case, so I think that this is not a valid solution at all." $\endgroup$ – Billel Hacaine Mar 7 '17 at 13:49
23
$\begingroup$

The problems are meant to be solved by sequential trials beginning in the ones place and noticing the ramifications of carry as the student follows a logical progression and tests the results of each step against the "rules" or requirements of the problem.

One "standard" rule that would be known to the students but is not obvious to a random adult is that within a problem, each letter represents a different digit. It is unfortunate that this is not explicitly stated, but the truth is that nearly any randomly selected homework problem from grade school will depend on a local understanding of what constraints may be assumed and what approaches are to be used.

For problem C we have ABC + ABC + ABC = BBB

The student is expected to notice that 
all the digits of the answer must be the same, and that 
the middle digit of abc must match all the digits of bbb.

Third grade thinking for Problem C proceeds like this:

Try C = 1

1 + 1 + 1 = 3, so b would equal 3. For the tens place, that means 3 + 3 + 3 would also need to have 3 in the digits place, but 3 + 3 + 3 = 9 so C cannot be 1.

Try C = 2

2 + 2 + 2 = 6, but 6 + 6 + 6 is 18, which can't be right because the digits place has to be 6, because all three digits of the answer have to be the same.

Try C = 3

3 + 3 + 3 = 9, but 9 + 9 + 9 = 27 when the ones place would have to be 9 to fit, so that's wrong too.

Try C = 4

4 + 4 + 4 = 12, so now B would have to be 2 and we need to carry when we test the tens digit: 2 + 2 + 2 (+ 1) = 7, which doesn't match 2, so wrong again.

Try C = 5

5 + 5 + 5 = 15, so B would be 5 and there's another carry: 5 + 5 + 5 (+ 1) = 16, which is wrong because we need the ones place to be 5.

Try C = 6

6 + 6 + 6 = 18, and 8 + 8 + 8 (+ 1) = 25, but 5 is not 8 so this is not the answer.

Try C = 7

7 + 7 + 7 = 21, so now we have to carry 2, and if B = 1 then 1 + 1 + 1 (+ 2) = 5, but we needed it to be 1, so this is wrong.

Try C = 8

8 + 8 + 8 = 24, so carry 2 again and try B = 4. Then 4 + 4 + 4 (+ 2) = 14. Hey! It matched! So then for the hundreds place we have to carry 1, so we need A to be a number that adds up to 4, and luckily 1 + 1 + 1 (+ 1) = 4, so YAY! This is correct!

Just to be sure, also try C = 9.

9 + 9 + 9 = 27, so carry 2 and B would be 7. Then 7 + 7 + 7 (+2) = 23, and this is wrong because 3 is not equal to 7.

At a higher grade level, students would be expected to notice that

3(abc) = bbb
thus
(bbb) / 3 = abc
and the middle digit of abc must be the same as all the digits of bbb.

Then the sequential trials step through all the possible values of bbb, and the thinking goes like this:

111 / 3 = 37 => too few digits

222 / 3 = 74 => too few digits

333 / 3 = 111 => wrong because 1 != 3

444 / 3 = 148 => correct!

555 / 3 => 185 => wrong because 8 != 5

666 / 3 = 222 => wrong because 2 != 6

777 / 3 = 259 => wrong because 5 != 7

888 / 3 = 296 => wrong because 9 !=> 8

999 / 3 = 333 => wrong because 3 != 9

= = = = =

Similarly, for problem B we have KLM + KLM = NKL.

EDIT: Thanks to those who pointed out that I had made an incorrect snap assumption that the final sum would increase monotonically with M; that of course is not true. I have now expanded the series. A third grader obviously would not make such a mistake. :)

The student should notice that when adding M + M, 
the ones place digit becomes L. Then that same L 
gets plugged into the tens place for the next column of addition, 
so the answer to L + L has a tens place digit that becomes K.

The student also must notice that the first two digits of KLM 
must match the last two digits of NKL. 

The student also must notice that although the ones place and the tens place
can have answers larger than 9 (resulting in a carry), the hundreds place      
must sum to less than ten, or else N would represent TWO digits, 
rather than one.  Another way of saying this would be that NKL can be 
up to 999 but no larger.

Here's how the thinking proceeds:

Try M = 1

1 + 1 = 2, so L would be 2. Then 2 + 2 = 4, so K would be 4. Then 4 + 4 = 8, so N is 8. 421 + 421 = 842. The first two digits of 421 DO match the last two digits of 842, so this is correct.

Try M = 2

2 + 2 = 4, so L is 4. Then 4 + 4 = 8, so K is 8. Then 8 + 8 = 16, so N would be 16. 842 + 842 = 1684. The first two digits of 842 do match the last two digits of 1684, but N can't be higher than 9 or else it would not be a digit, so this is wrong.

Try M = 3

3 + 3 = 6, so L would be 6. Then 6 + 6 = 12, so K would be 2. Then 2 + 2 (+1) = 5, so N = 5. Now we have 263 + 263 = 526. The first two digits of 263 match the last two digits of 526, so this is correct.

Try M = 4

4 + 4 = 8, so L = 8. Then 8 + 8 = 16, so K = 6. Then 6 + 6 (+ 1) = 13, but N can't be greater than 9, so this has to be wrong.

Try M = 5

5 + 5 = 10 so L = 0. Then 0 + 0 (+ 1) = 1, so K = 1. Then N = 1 + 1 = 2. We have 105 + 105 = 210, and the first two numbers of KLM do match the last two numbers of NKL, so this is correct.

Try M = 6

6 + 6 = 12, so L = 2. Then 2 + 2 (+ 1) = 5, so K = 5. Then 5 + 5 = 10, but of course N can't be bigger than 9 so this is wrong.

Try M = 7

7 + 7 = 14, so L = 4. Then 4 + 4 (+ 1) = 9, so K = 9. Now that I'm getting the hang of this I can see that if K is 5 or more then N will be greater than the maximum digit of 9, so I can say this is wrong even without calculating that N = 9 + 9 = 18.

Try M = 7

7 + 7 = 14, so L = 4. Then 4 + 4 (+ 1) = 9, so K = 9. That means N will be 18 -- too big -- so this is wrong (even though the first two digits of KLM do match the last two digits of NKL).

Try M = 8

8 + 8 = 16, so L = 6. Then 6 + 6 (+ 1) = 13, so K = 3. Then 3 + 3 (+ 1) = 7, so N is 7. We have 368 + 368 = 736. The first two digits of KLM do match the last two digits of NKL, so this is correct.

Try M = 9

9 + 9 = 18, so L = 8. Then 8 + 8 (+ 1) = 17, so K = 7, and 7 + 7 (+ 1) = 15, which is more than a single digit, so this is wrong.

Try M = 10. Oops, M can't be larger than 9 because it wouldn't be a digit, so we are done.

Isn't it interesting that even though some of the answers were too big for N to squeeze in with the digits, it was still possible to make this pattern work for every value of M from zero to nine?

= = =

Problem A is DEF + FEF = GHH

This one bothers adults because it does not yield a single unique solution, but that won't bother most third grade students. Full credit would be given for any correct answer. This example demonstrates that the student can notice the odd-even implications of carry and the fact that carry from adding in the hundreds place will give too many digits.

The student figures out that F + F gives an answer where 
the ones place digit is H. The observation that H in the digits place 
will always be an even number, while H in the tens place can only be 
an even number if there is no carry from the digits place (thus forcing
F to be less than 5) may be made at the outset or it may be discovered 
along the way. 

"F must be small enough that when you add F + F there's no carry, 
because when you add E + E in the tens place it will also be 
an even number, and if you then had to add a carried one, 
you'd get an odd number, and then that couldn't match 
the H in the ones place, which is even."

If this observation is not made, the student typically begins at the bottom       
and works her way upwards. if it *is* made at the outset, the student 
typically begins at 4 and works her way downward.

Third grade thought would be something like this:

F has to be less than 5, because 5 + 5 = 10 and that gives a carry digit, and I already know that since the tens place is E + E, I can't also have a carry digit and end up with an even number.

Try F = 4

4 + 4 = 8, so H is 8.

Now E cannot also be 4, since each letter in this problem represents a different digit. That means E has to be 9 because 9 + 9 = 18.

Now I have to carry one over to the hundreds place, and I have to put 4 in for F in the hundreds place.

That means before I even put D in I already have 4 (+1) = 5,

so I have room to let D be anything less than 5 (because if D were 5 or larger, the hundreds place wouldn't be a single digit).

But D cannot be 4 because that's already taken for F, so D can be 1, 2, or 3.

But if D were 3 then we'd have G = 3 + 4 (+1) = 8, and H is already 8, so D can be 1 or 2, making G = 6 or 7.

So the answers (FHEDG) for F = 4 are 48916 and 48927

= = =

Now try F = 3

3 + 3 = 6, so H would be 6 and E + E would have to be 16, so E must be 8.

8 + 8 is 16 so there's a carry to the hundreds place.

I already know F is 3, and 3 (+1) = 4, so D can be anything from 0 to 5 without causing an overflow.

But D cannot be 3 because it's already used for F, so D can be 1, 2, 4, 5.

But if D were 2, G would be 6, and 6 is already used by H. And if D were 4, G would be 8, which is already used by E.

So D can be 1 or 5, which makes G = 5 or 9.

So the answers (FHEDG) for F = 3 are 36815 and 36859

= = =

Try F = 2

2 + 2 = 4 so H = 4. E can't be 2 because F is 2, so E must be 7 because 7 + 7 > = 14. That means there's a carry to the hundreds place.

Putting in 2 for F, we have 2 (+1) = 3 before we add D, so D can be anything up to 6 without giving too many digits.

But D cannot be 2 or 4 because they're already used, so D can be 1, 3, or 5.

But if D = 1 then G = 4, which is taken, so D can be 3 or 5.

So the answers (FHEDG) for F = 2 are 24736 and 24758

= = =

Try F = 1

1 + 1 = 2, so H must be 2. That means E must be 6 because 6 + 6 = 12. [Hey, E was 8, 7, 6 when F was 4, 3, 2!]

Since we have a carry, when we put 1 in for F in the hundreds place we get 1 (+1) = 2, which means D could be any number up to 7 without causing an overflow.

But D cannot be 1 or 2 or 6, because those are already taken, so D can be 3, 4, 5, or 7.

But if D is 4 then G is 6, which is taken by E, so D can be 3, 5, or 7

So the answers (FHEDG) for F = 1 are 12635, 12657, 12679

= = =

Try F = 0

0 + 0 = 0, and that's no good because F and H can't be the same. In fact, if F = 0 then everything has to be zero, which shows that zero isn't at all like other numbers.

So yes, it does make sense, and it can be solved by a third-grader who has been taught this approach to add and carry, and yes there are many other ways to approach this kind of problem.

In this context it's all about add and carry, and about noticing the impact of carry on the next column.

I give the results of Problem A as FHEDG because that's the order in which the child solves the problem.

I am new at this site and do not have the rep to comment, so I'll point out here that Phil's interesting perl solution doesn't fully apply the "standard rule" that for a given problem, each letter represents a different digit. Phil has applied that rule for D, E, F, and H, but has relaxed it for G, which seems a bit arbitrary. If multiple letters can represent the same number, we could also have things like FHEDG = 36336, right?

$\endgroup$
  • 1
    $\begingroup$ You know, the thing about that standard rule is that it's not actually included in the information given to the child. Unfortunate, that. $\endgroup$ – the dark wanderer Mar 6 '17 at 6:55
  • 1
    $\begingroup$ In problem (b): What do you mean by "And if that answer is too big, then any larger numbers must also be too big, so we can stop right here!!"? $\endgroup$ – Billel Hacaine Mar 6 '17 at 10:25
  • $\begingroup$ To @BillelHacaine's point, in the second problem M can be bigger than 2. You are ignoring the ability to sum two digits to a number greater than 10. Try M = 5 L = 0 K = 1 N = 2 as another solution to KLM + KLM = NKL. $\endgroup$ – Mark Balhoff Mar 8 '17 at 1:49
  • $\begingroup$ Thanks to Hacaine and Balhoff who pointed out my error in B; I have edited to correct this. I also appreciate the editor who corrected my arithmetic where I confusedly added two values when the formula required 3. My point was just that this kind of problem is not beyond the kind of thinking we expect from a third grader, if we have previously taught them how to approach such substitutions. This is obviously part of a unit on addition with carry, and I think it's actually pretty cool as homework goes... $\endgroup$ – Craig.Feied Mar 9 '17 at 8:44
13
$\begingroup$

A good place to start would be with $ABC + ABC + ABC = BBB$. Since $BBB$ can only take one of $6$ values (it must be at least $123+123+123 = 369$ so it's one of $444,555,\ldots,999$), $ABC$ is uniquely determined as one-third of $BBB$. The fact that $B$ appears in both numbers narrows down the field even further, as shown in Billel Hacaine's answer: $A=1, B=4, C=8$.

From $DEF + FEF = GHH$ we have from the one's digit that $H$ is even, hence there is no carry and thus $2F = H$. Since the last digit of $2E$ is also $H$ but $F \ne E$, it follows that $E = F+5$. There us no choice of $F$ for which $\{F,F+5,2F\}$ is disjoint from $\{1,4,8\}$, though.

$\endgroup$
  • 2
    $\begingroup$ Why is F != E? that is stated nowhere, or do I miss something? $\endgroup$ – Jens Schauder Mar 5 '17 at 6:44
  • 4
    $\begingroup$ @Charles H is even, 2E is even too. Therefore, H can't be equal to 2E + 1. (And the carry can be at most 1) $\endgroup$ – Ken Bourassa Mar 5 '17 at 6:50
  • 8
    $\begingroup$ @JensSchauder That is an automatic assumption in verbal arithmetic and the children would have been told of it before seeing the homework. $\endgroup$ – The Vee Mar 5 '17 at 9:14
  • 3
    $\begingroup$ @TheVee: I know that. But why should we assume that $E \ne F$ when we don't assume that $A \ne E$, and all other pairs of letters as well? If (a), (b) and (c) are supposed to be considered completely separate questions, it makes no sense to use different letters in all of them. If they are supposed to be considered one question, then there are more letters than digits so it makes no sense to assume that different letters cannot represent the same digit. If your assumption is right then this is a very poorly described question. $\endgroup$ – user21820 Mar 5 '17 at 16:06
  • 5
    $\begingroup$ @user21820 "it makes no sense to use different letters in all of them" It makes perfect sense to use different letters to more easily separate "the A in (a)" from "the A in (c)" when filling in the answers. While it would have been better to explicitly state "in each problem the letters are different" but as Billel says, we need to "think as a kid but not as an advanced expert", and they'd probably just spent the day doing similar problems with similar conditions. $\endgroup$ – TripeHound Mar 6 '17 at 10:34
11
$\begingroup$

There are different ways to solve this.

Given the age and school year, I think a logical trial and error approach is what they expect.

Assume that any number is possible for any character, then try them to see what follows. Knowing basic addition, the possibilities are far fewer than it may seem at first.

I'm starting with the ABC problem, since it looks to me like the easiest one (many repetitions, only 3 letters).

Since addition starts with the smallest value (C) that's where we begin testing numbers.

If C=0 then since C+C+C=B we get B=0, giving

 A00
+A00
+A00
----
 000

The only possibility here is A=0, and this is a valid solution but too simple so I'd say we can ignore it and look for another.

If C=1 then B=3 and we get

 A31
+A31
+A31
----
 333

The second column isn't correct since 3+3+3=9 so it's safe to say C is not 1.

Continue like this.

C=2 => B=6 same problem

C=3 => B=9 same problem

C=4 => B=2 (now we have a 1 carry) which gives for column 2: 1+2+2+2=2 , also wrong

You'll find that the only possibility is C=8 which gives you

 12
 A48
+A48
+A48
----
 444

now you solve for A which is 1 since 1+A+A+A=4

For the KLM problem try 1 for M giving

 421
+421
----
 842

Or M=3 giving

  1
  263
 +263
 ----
  526

Or M=5 giving

   1
  105
 +105
 ----
  210

There are a few possibilities here (I didnt try all of them). If M = 2 or 4 on the other hand, you end up with a carry in the first column giving you a 4 digit sum, which isn't the case, so these are not options.

For the DEF problem, assuming that different letters represent different numbers:

One solution is

 1     
 361
+161
----
 522
$\endgroup$
7
$\begingroup$

For the last equation, $3(100A+10B+C) = 111B$ and hence $100A+C = 27B$. Since $0 \le C \le 9$, you basically need a multiple of $27$ that is very close to a multiple of $100$. Only $27 \times 4$ works, so you know $A,B,C$.

For the second equation you just work from right to left, guessing $M$ and finding $L,K,N$ in order. For example if you guess that $M = 1$, you get that $L = 2$ and $K = 4$ and $N = 8$. But there is more than one solution, actually four in total!

For the first equation, nearly anything goes. We cannot impose that each letter stands for a unique digit, because there are too many letters.

$\endgroup$
  • $\begingroup$ +1 for knowing what $ABC$ in decimal system really means. $\endgroup$ – Jagger Mar 6 '17 at 11:27
  • 1
    $\begingroup$ @Jagger: Yes it is a pity that all the top-voted answers are just whacking with brute-force trial-and-error. $\endgroup$ – user21820 Mar 6 '17 at 11:29
  • $\begingroup$ I posted three comments with solutions under the original question. If you don't mind adding them to your answer or creating a new one, please do. Apparently I must have 10 points to post an answer here, and 100 hundred welcome points do not count. :( $\endgroup$ – Jagger Mar 6 '17 at 11:59
6
$\begingroup$

The teacher did not see fit to write "each letter stands for a different digit."

Therefore, "all letters are equal to the number zero" is a valid answer, and should earn maximum points. I would argue that it should earn above maximum, since it also has higher educational value, and homework is all about educational value.

Regarding math, this answer provides an important lesson: the solution to a problem sometimes isn't the most obvious one. Sometimes, when making a proof, a non-obvious approach yields an easier result.

Regarding engineering, this answer provides an even more important lesson: sometimes, the best solution to a problem is to get rid of the problem.

Having to argue why this answer is valid also has educational value.

I think this rather pointless homework provides a wonderful opportunity to teach her several life-saving skills, that will help her immensely in all stages of her life:

  • Lateral thinking (ie, out of the box).
  • Critical thinking (ie, question blind trust in authority)
  • And, last but not least, the subtle and holy art of Trolling.

Teaching her at an early age that every word matters will be of immense help when she later has to parse lawyer language in contracts.

Critical thinking means every word written on a piece of paper might not be true. In fact, considering the state of our press, one should never read it without one's bullshit filter set to maximum.

There is also a hidden bonus. When the teacher inevitably grades her paper as FAIL, because most teachers are idiots, your dad will get to file complaint, and have fun in the director's office.

And this will earn him (and you) her unconditional respect.

$\endgroup$
  • 3
    $\begingroup$ As noted in comments elsewhere, the digits shouldn't all be zero because that's improper notation. The individual numbers must all be three digit numbers. It's good to be clever, but not at the expense of being unable to understand what other people actually mean. $\endgroup$ – Era Mar 6 '17 at 19:23
5
$\begingroup$

I found by far the easiest way to solve (a) was thru brute force Perl command line:

perl -e "for ($dd=1;$dd<10;$dd++) { for ($ee=0;$ee<10;$ee++) { for ($ff=1;$ff<10;$ff++) { for ($gg=1;$gg<10;$gg++) { for ($hh=0;$hh<10;$hh++) { next if $ee == $ff; next if $dd == $ee; next if $dd == $ff; next if $gg == $hh; $result=qq/$dd$ee$ff/+qq/$ff$ee$ff/; next if $result == 0; if ($result == qq/$gg$hh$hh/) { print qq/$ff $hh $ee $gg $dd\n/; }}}}}} " | sort

Note: this was using ActiveState.com's community version of Perl v5.16.2 (last version that there's a working module for Image::Magick on bribes.org)

Results of the above command line:

1 2 6 4 2

1 2 6 5 3

1 2 6 6 4

1 2 6 7 5

1 2 6 9 7

2 4 7 6 3

2 4 7 7 4

2 4 7 8 5

2 4 7 9 6

3 6 8 5 1

3 6 8 8 4

3 6 8 9 5

4 8 9 6 1

4 8 9 7 2

P.P.S. Note that the output is in F H E G D order as I was comparing results to the answer above mine =)

$\endgroup$
  • 7
    $\begingroup$ I would love to see the teacher's face when the pupil writes down the Perl program as answer. :D $\endgroup$ – AnoE Mar 6 '17 at 7:55
4
$\begingroup$

For the 3rd sum:Let the carry from the 1st column to the 2nd column be $d$, and let the carry from the 2nd column to the 3rd column be $e$.

The second column implies that $3B+d $ is equal to $B$ plus a multiple of $10,$ so $2B+d$ is a multiple of $10$. So $d$ is even, so $d=0$ or $d=2$.

If $d=0$ then $2B=2B+d$ is a multiple of $10$ so $B=0$ or $B=5$. Now $B=0$ gives the "trivial" solution $A=B=C=0$ (because $3\cdot ABC=BBB=000=0$.) But if $d=0$ and $B=5$ then $3C=B+10d=B=5$, which is impossible.

So a no-trivial solution must have $d=2.$ The $2B+2=2B+d$ is a multiple of $10$, which requires $B=4$ or $B=9$.

But if $d=2$ and $B=9$ then $3C=10d+B=20+9=29,$ which is impossible.

With $d=2$ and $B=4$ we have $3C=10d+B=20+4=24$, giving $C=8.$ Finally $3B+d=12+2=14$, giving the carry $e=1.$ And $3A+1=3A+e=B=4,$ giving $A=1$. (Solution:$148+148+148=444$).

8-year-olds who can solve this on their own would be rare.

$\endgroup$
4
$\begingroup$

The homework is badly written. It should tell you that each letter represents a different number from 0-9 and you are supposed to figure out what numbers the letters mean. It's supposed to be a math problem about thinking about what 'carrying the one' means when two numbers sum to larger than 10. Because the problem is poorly worded your child is not well-equipped to solve the problem on her own (there is no particular reason from the perspective of a 8-year-old child why that is a good rule and not, e.g. the alphanumeric rule you postulated or some variety of word game e.g. DEF +FEF = GHH, GHH is the sound one makes when frustrated, so the problem is about how you can't add words, only numbers).

When my younger sister struggles with such a problem, I review the question to ensure there's no instructions she missed that would make the problem clear, then ask her to write a short explanation as to why the problem doesn't make sense instead of answering it (which I hope will prepare her for higher math where proving that the question asked is impossible is an expected approach for certain kinds of problems). She gets practice thinking about math and how to interpret different kinds of questions and tell what part of something she's doing is the 'hard' part, which I think more than makes up for whatever she's 'supposed' to be getting out of such poorly worded questions.

My sister is now in 8th grade, and asking her to write down what makes things hard in 3rd grade would probably not have gone as well. Writing is hard for a 3rd grader! When she was younger, we would just talk about what made the problem hard for her and I would try and get her just to get past 'I/you can't do it/don't know how to do it' (the immediate answer of any 3rd grader told to do something hard) and onto the reasons why she felt she couldn't do it. I'd then either show her what I thought the teacher wanted her to do or just have her leave it blank and tell her it's ok not to be able to do the problems as long as you try.

Be aware that figuring out why something is hard is itself really hard sometimes, especially for children, and especially the first several times they are asked to do it. You'll probably spend at least 30 minutes on that question and you'll need to spend a significant part being quiet-but-supportive/understanding while your child thinks.

$\endgroup$
  • 1
    $\begingroup$ Homework is not given totally bare of context. It is almost guaranteed that the class has just finished a subtopic on column addition and carriage of surplus. The problem makes perfect sense when one is aware of how the previous lessons (have probably) progressed. $\endgroup$ – Nij Mar 6 '17 at 6:01
  • 2
    $\begingroup$ @Nij no, it doesn't. It's true that the class (hopefully) should have just covered the topic but problems should be phrased clearly, especially for children. Nowhere (that the eight year old, her mother, or I can see) does the assignment explain that the letters in each problem are different numbers, but the letters in later problems might be the same numbers as letters in an earlier one. This is really important when introducing variables to someone who hasn't been taught anything about variables yet. No where does it say that there are multiple possible solutions. $\endgroup$ – the dark wanderer Mar 6 '17 at 6:26
  • $\begingroup$ It doesn't have to. The fact there are multiple solutions does not disallow one solution from being correct even though others exist, nor is there a need to assume all letters are different in the first place. The latter fact can be discerned logically just from making an attempt, and the former should only be discerned by a student who has done enough work to discover it, therefore implicitly have the capacity to understand and work with it. There are 15-year-olds who freak out when told there is more than one correct solution; expecting 8-year-olds to handle it better is not justified. $\endgroup$ – Nij Mar 6 '17 at 6:36
  • 1
    $\begingroup$ @Nij But really the real solution here is to not give third graders ambiguous math problems in the first place. Or to teach them about that especially. But definitely not this. $\endgroup$ – the dark wanderer Mar 6 '17 at 6:52
  • 1
    $\begingroup$ I am a teacher and have taught all levels from kindergarten to college. It is IMPOSSIBLE to give an unambiguous math question. By the time you have covered every possible question of definition and possibilities and alternative theories, the school year would be long over. Eight year olds in Grade 3 do NOT need or want detailed explanations of every possible different interpretation. After two or three their eyes glaze over, and after five or six you might be able to make them quit math for life. You have to pitch the question to the student. Properly, puzzle problems are taught by example. $\endgroup$ – victoria Mar 10 '17 at 5:09
4
$\begingroup$

There is no unique answer to (a). The intention must be to expose grade school students to normal adult frustrations. To find all those answers start with F. Other answers have shown that it must be 1,2,3 or 4. For each F there is a unique H (2,4,6,8) and E (6,7,8,9), respectively. Again, the other answers get that far. Then G can be any number that isn't F or H or E. There is a unique D for each G, but some of them are 0 (criptarithms are understood to not have leading 0s) and others are negative. The solutions are:

F H E G D

1 2 6 5 3

1 2 6 7 5

1 2 6 9 7

2 4 7 6 3

2 4 7 8 5

2 4 7 9 6

3 6 8 5 1

3 6 8 9 5

4 8 9 6 1

4 8 9 7 2

I used a spreadsheet.

I thought the answer to (b) was easy to get by inspection. Several answers have the same answer as me to (c): A,B,C are 1,4,8.

$\endgroup$
  • 1
    $\begingroup$ What about the ones where $F=3$? I get two of them (with $D$ to $H$ all distinct). And of course $G$ cannot be just any number, but must be $ > F+1 $ $\endgroup$ – ilkkachu Mar 5 '17 at 21:46
  • 1
    $\begingroup$ @ilkkachu I thought I posted a comment agreeing with yours, but I can't find it now. I edited accordingly. The F=3 solutions didn't make it from my spreadsheet to my answer when I copied. $\endgroup$ – stretch Mar 11 '17 at 19:20
4
$\begingroup$

The easiest way for an 8 year old would be to try all combinations until one fits. Taking $F=1$ in a) and $M=1$ in b) gives an immediate answer, but only until checking upto $C=8$ do we find an answer for c). Discounting the trivial case a) has 48 possible answers, b) has 4 and c) has 1.

a)

$F=0$ is trivial.

$F=1$ has $2F=H=2$ and $E=1$ or $E=6$. For $E=1$, then $D=1,2,3,4,5,6,7,8$ gives $D+F=G=2,3,4,5,6,7,8,9$ as possible answers, and for $E=6$, $D=1,2,3,4,5,6,7$ gives $D+F+1=G=3,4,5,6,7,8,9$ as possible answers.

$F=2$ has $2F=H=4$ and $E=2$ or $E=7$. For $E=2$, then $D=1,2,3,4,5,6,7$ gives $D+F=G=3,4,5,6,7,8,9$ as possible answers, and for $E=7$, $D=1,2,3,4,5,6$ gives $D+F+1=G=4,5,6,7,8,9$ as possible answers.

$F=3$ has $2F=H=6$ and $E=3$ or $E=8$. For $E=3$, then $D=1,2,3,4,5,6$ gives $D+F=G=4,5,6,7,8,9$ as possible answers, and for $E=8$, $D=1,2,3,4,5$ gives $D+F+1=G=5,6,7,8,9$ as possible answers.

$F=4$ has $2F=H=8$ and $E=4$ or $E=9$. For $E=4$, then $D=1,2,3,4,5$ gives $D+F=G=5,6,7,8,9$ as possible answers, and for $E=8$, $D=1,2,3,4$ gives $D+F+1=G=6,7,8,9$ as possible answers.

$F=5$ has $2F=10$, $H=0$ and the unit part of $2E$ is $H+1=1$, so this case cannot occur.

$F=6$ has $2F=12$, $H=2$ and the unit part of $2E$ is $H+1=3$, so this case cannot occur.

$F=7$ has $2F=14$, $H=4$ and the unit part of $2E$ is $H+1=5$, so this case cannot occur.

$F=8$ has $2F=16$, $H=6$ and the unit part of $2E$ is $H+1=7$, so this case cannot occur.

$F=9$ has $2F=18$, $H=8$ and the unit part of $2E$ is $H+1=9$, so this case cannot occur.

b)

$M=0$ is trivial.

$M=1$ has $2M=2$, then $L=2$ and $2L=K=4$ and $2K=8=N$ so this is a possible answer.

$M=2$ has $2M=4$, then $L=4$ and $2L=K=8$ and $2K=16=N$ but $16$ is two digits so this case cannot occur.

$M=3$ has $2M=6$, then $L=6$ and $2L=12$ and so $K=2$ and $2K=4=N$ so this is a possible answer.

$M=4$ has $2M=8$, then $L=8$ and $2L=16$ and so $K=6$ and $2K=12=N$ but $12$ is two digits so this case cannot occur.

$M=5$ has $2M=10$, then $L=0$ and $2L=0$ and so $K=0+1$ and $2K=2=N$ so this is a possible answer.

$M=6$ has $2M=12$, then $L=2$ and $2L=4$ and so $K=4+1=5$ and $2K=10=N$ but $10$ is two digits so this case cannot occur.

$M=7$ has $2M=14$, then $L=4$ and $2L=8$ and so $K=8+1=9$ and $2K=18=N$ but $18$ is two digits so this case cannot occur.

$M=8$ has $2M=16$, then $L=6$ and $2L=12$ and so $10+K=12+1=13$ so $K=3$ and $2K+1=7=N$ so this is a possible answer.

$M=9$ has $2M=18$, then $L=8$ and $2L=16$ and so $K=6+1=7$ and $2K+1=15=N$ but $15$ is two digits so this case cannot occur.

c)

$C=0$ is trivial.

$C=1$ has $3C=3$, then $B=3$ and $3B=9\neq3$ so this case cannot occur.

$C=2$ has $3C=6$, then $B=6$ and $3B=10+8$ but $8\neq6$ so this case cannot occur.

$C=3$ has $3C=9$, then $B=9$ and $3B=20+7$ but $7\neq9$ so this case cannot occur.

$C=4$ has $3C=10+2$, then $B=2$ and $3B+1=6+1=7$ but $7\neq2$ so this case cannot occur.

$C=5$ has $3C=10+5$, then $B=5$ and $3B+1=10+5+1=10+6$ but $6\neq5$ so this case cannot occur.

$C=6$ has $3C=10+8$, then $B=8$ and $3B+1=20+4+1=20+5$ but $5\neq8$ so this case cannot occur.

$C=7$ has $3C=20+1$, then $B=1$ and $3B+2=3+2=5$ but $5\neq1$ so this case cannot occur.

$C=8$ has $3C=20+4$, then $B=4$ and $3B+2=12+2=10+4$ and $B=4$, then finally $3A+1=4$ and so $A=1$ so this case is an answer (the only one).

$C=9$ has $3C=20+7$, then $B=7$ and $3B+2=21+2=20+3$ but $7\neq3$ so this case cannot occur.

$\endgroup$
3
$\begingroup$

It is clear that problems a), b), c) are independent each other.

I find just solutions maybe not in an exhaustive way.

a) DEF+FEF=GHH

$DEF+FEF=GHH\implies F+F=H \text { or }1H$

$F+F=H\implies (F,H)\in\{(0,0),(1,2),(2,4),(3,6),(4,8)\}$

$F+F=1H\implies(F,H)\in\{(5,0),(6,2),(7,4),(8,6),(9,8)\}$

We discard the case $(F,H)=(0,0)$ because it gives the "solutions" $100+000=100$ and $150+050=200$ and we don't consider $000$ nor $050$.

Now $(F,H)=(1,2)$ gives $DE1+1E1=G22\implies E=1$ and $D+1=G$ which gives the solutions $(D,G)\in\{(1,2),(2,3),(3,4),(4,5),(5,6),(6,7),(7,8),(8,9)\}$.

We have in this case the sum $D11+111=G22$ with the consequent eight given values for $(D,G)$.

b) KLM+KLM=NKL

$2*(KLM)=NKL\implies L\in\{0,2,4,6,8\}$ $L=0\implies M=5\text{ or } 0\implies 2*(K05)=N10\implies K=1\implies N=2$.

(the value $M=0$ gives discarded trivial solution).

We have the sum $105+105=210$ as a solution.

c) ABC+ABC+ABC=BBB

We have $3*(ABC)=111*B\implies ABC=37*B\implies B\gt 2\implies B=4$ because for $B\in\{3,5,6,7,8,9\}$ we would have $37*B=XYZ$ with $Y\ne B$.

Now $3*A4C=444\implies A4C=148\implies (A,C)=(1,8)$

In this case we have $148+148+148=444$.

$\endgroup$
2
$\begingroup$

I am assuming that we have three separate problems, we are in number base 10, that there are no leading zeroes, and that there is a one-to-one correspondence between digits and letters.

For the first problem, observe that H is even because F+F=H. Also note that F+F does not generate a carry, because E+E is also even. So there are four possibilities for (F,H,E): (1,2,6),(2,4,7),(3,6,8),(4,8,9). Each of these has several possibilities for D and G, leasing to 13 solutions for (F,H,E,D,G): (1,2,6,3,5), (1,2,6,5,7), (1,2,6,7,9); (2,4,7,0,3), (2,4,7,3,6), (2,4,7,5,8), (2,4,7,6,9); (3,6,8,0,4), (3,6,8,1,5), (3,6,8,4,7); (4,8,9,0,5), (4,8,9,1,6), (4,8,9,2,7). If we assume that the solution is unique, other number bases, starting with 5, could be explored.

For second problem, there are four possibilities for K because K+K does not generate a carry. Also, L must be even. An even guess for K rules out a carry for M+M, while an odd guess for K requires a carry for M+M. This greatly limits the possibilities to consider. There are four possible values for K, and the solutions (K,L,M,N) are (1,0,5,2), (2,6,3,5), (3,6,8,7), (4,2,1,8), and (4,6,3,9).

Four the third problem, note that there are nine possible numbers of the form BBB. But one third of BBB is a three digit number only for B>2. If you divide each of the three digit numbers 333, 444, 555, etc. by 3 only one of them, 444, gives a quotient with the same middle digit. So (A,B,C)=(1,4,8).

A great book that might help with this is beyond a third grade student, but might help us seniors with Altzheimer's: Doerfler's Dead Reckoning.

Another method of solution would be a generate and test procedure on a programmable calculator. My favorite would be the WP-34s. Although it runs on a discontinued platform, you can emulate it on your iPhone.

What? A third grade girl with no iPhone? That is child abuse!!!

$\endgroup$
1
$\begingroup$

Since there seems to be no answer to the second equation yet I will have a go at it:

For this is a third grade task I will be assuming that all variables have positive natural numbers as values.

We can see that $N$ is a single digit. Therefore we know that $N < 10$. We also know that $2K = N$. If we substitute the $N$ in the first equation for $2K$ and devide the whole formula by $2$ we get $K < 5$.

The other two digits can now be determined similarly: $K < 5$ and $K = 2L$, therefore $L < 2.5$. Since we are working with natural numbers this is equivalent to $L \le 2$.

$L \le 2$ and $L = 2M$, therefore $M \le 1$.

So $M$ can either be $0$ or $1$. However, if $M$ was $0$, the equation $2M = L$ would have to be $2M = M$ instead (I am quity sure that a third grade exercise won't have two Variables with the same value). That leaves us with only one conclusion: $M = 1$. With that knowledge calculating the other variables is easy:

$M=1$
$L=2M = 2$
$K = 2L = 4$
$N = 2K = 8$

$\endgroup$
  • $\begingroup$ This is a totally false answer. Why is it even upvoted? $\endgroup$ – user21820 Mar 10 '17 at 6:56

protected by Community Mar 6 '17 at 7:18

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.