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How many $3$-digit numbers can be formed so that the sum of two digits will be equal to the third digit?

I am confused in this question whether to take first 2 digits sum or any digit sum such that it is equal to 3rd digit.

For example, some combinations 112 can be written into 3 ways 112 112 121 such that in the last

Example 3rd number and first number sum = 2 (middle number).

Also, is there any better approach towards this problem rather than counting and arranging the 3 digit number?

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    $\begingroup$ Your confusion is completely reasonable, the problem as quoted is not well-stated and there are at least two reasonable interpretations. $\endgroup$ – André Nicolas Aug 26 '15 at 14:38
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    $\begingroup$ Have you reproduced the question exactly ? "sum of the 2 digits" is not even proper English. $\endgroup$ – true blue anil Aug 26 '15 at 14:39
  • $\begingroup$ @trueblueanil reproduced the correct english form of the question. $\endgroup$ – justin takro Aug 26 '15 at 14:44
  • $\begingroup$ @AndréNicolas yes What to do i think Calculate and See the options side by side $\endgroup$ – justin takro Aug 26 '15 at 14:45
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    $\begingroup$ If you have reproduced, (not changed) to correct English form, I would interpret it as the sum of any 2 digits is equal to the third. $\endgroup$ – true blue anil Aug 26 '15 at 14:49
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I'm going to assume that the problem means that the sum of any two digits equals the third.

To start, let's find out how many 3 digit numbers have the last digit as the sum of the first two digits:

Here, the first digit cannot be $0$ or $9$. It can't be $9$ because the sum of $9$ and any digit is a two digit number If the first digit is $1$ then second number can be any digit from $0$ to $8$. So nine possibilities. If the first digit is $2$ then the possibities for the second digit is $0$ through $7$.

So, the number of three digit numbers whose first two digits add up to the third is: $9+8+\cdots +1$.

Now, the total of all numbers in which the sum any two digits is 3 times the previous answer because the sum digit could be the first, second or third digit.

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  • $\begingroup$ So,according to you Ans is 135 $\endgroup$ – justin takro Oct 10 '15 at 14:17
  • $\begingroup$ Yes. That's what I get. $\endgroup$ – Tim Raczkowski Oct 10 '15 at 14:21
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    $\begingroup$ Why can't the first digit be 9? 909 has as last digit the sum of the other digits. $\endgroup$ – Christian Sievers Jul 28 '16 at 12:27
  • $\begingroup$ Yes, you are correct. $\endgroup$ – Tim Raczkowski Aug 1 '16 at 1:26
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To write the problem in somewhat mathematical terms, we have three digits $d_1 \in \{1, \dots, 9\}$, $d_2,d_3 \in \{0, 1, \dots, 9\}$ that form a $3$-digit number $$x = d_1 \cdot 10^2 + d_2 \cdot 10 + d_3$$ Now, we want to find all possible values of $x$ such that at least one of the following holds:

  • $d_1 + d_2 = d_3$,
  • $d_1 + d_3 = d_2$,
  • $d_2 + d_3 = d_1$

Let's start with the first condition. We may pick $d_1$ freely, which gives us $9$ options. However, since $d_3 \in \{0, 1, \dots, 9\}$, we must always choose $d_2 \in \{0, \dots, 9-d_1\}$. Hence: $$\# \{ x \mid d_1 + d_2 = d_3\} = \sum_{d_i = 1}^{9}(10-d_i) = 45$$ The same reasoning holds for the second condition: $$\# \{ x \mid d_1 + d_3 = d_2\} = \sum_{d_i = 1}^{9}(10-d_i) = 45$$ However, the last condition is different! We pick $d_2$ freely which gives us $10$ options, and we must choose $d_3 \in \{0, \dots, 9-d_2\}$. However, we also have to exclude one case: $d_2 = d_3 = 0$, because then $d_1$ would also be $0$. Hence: $$\# \{ x \mid d_2 + d_3 = d_1\} = \sum_{d_i = 0}^{9}(10-d_i)-1 = 54$$ We now have $45 + 45 + 54 = 144$ numbers $x$ such that either of the conditions hold. However, some numbers are counted twice. It is easy to see that a number is counted twice if and only if $d_i = d_j$ and $d_k = 0$ for $i,j,k$ being some permutation of $1,2,3$. Since $d_1 \neq 0$, the multiples we need to discard are those of the form $d_2 = 0$ or $d_3 = 0$. There are $9$ of each ($d0d$ and $dd0$ with $d = 1, \dots, 9$), so we remove $18$ multiples. There is no number for which all $3$ conditions hold, since that would require $d_1 = d_2 = d_3 = 0$.

The solution is then: $$\#\{x\} = 144- 18 = 126.$$

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  • $\begingroup$ You might mention that there are no solutions where all three conditions hold, as you would have to add those back (priciple of inclusion and exclusion) $\endgroup$ – Christian Sievers Jul 28 '16 at 12:23
  • $\begingroup$ @ChristianSievers You're right, I've added that. $\endgroup$ – TastyRomeo Jul 28 '16 at 13:24
  • $\begingroup$ @DPoole Only the condition $d_1 + d_2 = d_3$ holds for that one. $\endgroup$ – TastyRomeo Jul 28 '16 at 13:24
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Answer is 126.

Where X is first digit (1 through to 9) and y is number of digits use equation below and add answers:

             y=  2(9-X) + ( 10-(9-X))

eg X=1 2(9-1) + ( 10 - (9-1)) =18 Then working through remaining first digits: X=2 y=17, X=3 y=16......X=9 y=10 Sum of the Y values is 126

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One answer says $126$, the other $135$. Which one is correct?

Brute-forcing in Haskell,

λ filter (\(a,b,c)->(a+b==c || a+c==b || b+c==a)) [ (x2,x1,x0) | x0 <- [0..9], x1 <- [0..9], x2 <- [1..9] ]
[(1,1,0),(2,2,0),(3,3,0),(4,4,0),(5,5,0),(6,6,0),(7,7,0),(8,8,0),(9,9,0),(1,0,1),(2,1,1),(1,2,1),(3,2,1),(2,3,1),(4,3,1),(3,4,1),(5,4,1),(4,5,1),(6,5,1),(5,6,1),(7,6,1),(6,7,1),(8,7,1),(7,8,1),(9,8,1),(8,9,1),(2,0,2),(1,1,2),(3,1,2),(4,2,2),(1,3,2),(5,3,2),(2,4,2),(6,4,2),(3,5,2),(7,5,2),(4,6,2),(8,6,2),(5,7,2),(9,7,2),(6,8,2),(7,9,2),(3,0,3),(2,1,3),(4,1,3),(1,2,3),(5,2,3),(6,3,3),(1,4,3),(7,4,3),(2,5,3),(8,5,3),(3,6,3),(9,6,3),(4,7,3),(5,8,3),(6,9,3),(4,0,4),(3,1,4),(5,1,4),(2,2,4),(6,2,4),(1,3,4),(7,3,4),(8,4,4),(1,5,4),(9,5,4),(2,6,4),(3,7,4),(4,8,4),(5,9,4),(5,0,5),(4,1,5),(6,1,5),(3,2,5),(7,2,5),(2,3,5),(8,3,5),(1,4,5),(9,4,5),(1,6,5),(2,7,5),(3,8,5),(4,9,5),(6,0,6),(5,1,6),(7,1,6),(4,2,6),(8,2,6),(3,3,6),(9,3,6),(2,4,6),(1,5,6),(1,7,6),(2,8,6),(3,9,6),(7,0,7),(6,1,7),(8,1,7),(5,2,7),(9,2,7),(4,3,7),(3,4,7),(2,5,7),(1,6,7),(1,8,7),(2,9,7),(8,0,8),(7,1,8),(9,1,8),(6,2,8),(5,3,8),(4,4,8),(3,5,8),(2,6,8),(1,7,8),(1,9,8),(9,0,9),(8,1,9),(7,2,9),(6,3,9),(5,4,9),(4,5,9),(3,6,9),(2,7,9),(1,8,9)]

λ length $ filter (\(a,b,c)->(a+b==c || a+c==b || b+c==a)) [ (x2,x1,x0) | x0 <- [0..9], x1 <- [0..9], x2 <- [1..9] ]
126

If $x_2 = 0$ is allowed, then

λ filter (\(a,b,c)->(a+b==c || a+c==b || b+c==a)) [ (x2,x1,x0) | x0 <- [0..9], x1 <- [0..9], x2 <- [0..9] ]
[(0,0,0),(1,1,0),(2,2,0),(3,3,0),(4,4,0),(5,5,0),(6,6,0),(7,7,0),(8,8,0),(9,9,0),(1,0,1),(0,1,1),(2,1,1),(1,2,1),(3,2,1),(2,3,1),(4,3,1),(3,4,1),(5,4,1),(4,5,1),(6,5,1),(5,6,1),(7,6,1),(6,7,1),(8,7,1),(7,8,1),(9,8,1),(8,9,1),(2,0,2),(1,1,2),(3,1,2),(0,2,2),(4,2,2),(1,3,2),(5,3,2),(2,4,2),(6,4,2),(3,5,2),(7,5,2),(4,6,2),(8,6,2),(5,7,2),(9,7,2),(6,8,2),(7,9,2),(3,0,3),(2,1,3),(4,1,3),(1,2,3),(5,2,3),(0,3,3),(6,3,3),(1,4,3),(7,4,3),(2,5,3),(8,5,3),(3,6,3),(9,6,3),(4,7,3),(5,8,3),(6,9,3),(4,0,4),(3,1,4),(5,1,4),(2,2,4),(6,2,4),(1,3,4),(7,3,4),(0,4,4),(8,4,4),(1,5,4),(9,5,4),(2,6,4),(3,7,4),(4,8,4),(5,9,4),(5,0,5),(4,1,5),(6,1,5),(3,2,5),(7,2,5),(2,3,5),(8,3,5),(1,4,5),(9,4,5),(0,5,5),(1,6,5),(2,7,5),(3,8,5),(4,9,5),(6,0,6),(5,1,6),(7,1,6),(4,2,6),(8,2,6),(3,3,6),(9,3,6),(2,4,6),(1,5,6),(0,6,6),(1,7,6),(2,8,6),(3,9,6),(7,0,7),(6,1,7),(8,1,7),(5,2,7),(9,2,7),(4,3,7),(3,4,7),(2,5,7),(1,6,7),(0,7,7),(1,8,7),(2,9,7),(8,0,8),(7,1,8),(9,1,8),(6,2,8),(5,3,8),(4,4,8),(3,5,8),(2,6,8),(1,7,8),(0,8,8),(1,9,8),(9,0,9),(8,1,9)

λ length $ filter (\(a,b,c)->(a+b==c || a+c==b || b+c==a)) [ (x2,x1,x0) | x0 <- [0..9], x1 <- [0..9], x2 <- [0..9] ]
136

Even if we remove $000$ from the list, which would give $135$ numbers, $011$ is not a $3$-digit number.

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    $\begingroup$ You're right, the answer is indeed $126$. There are $45$ numbers such that $d_1 + d_2 = d_3$, as the accepted answer says, but one cannot simply multiply this by $3$ and claim to have the answer, because the problem is not symmetric due to $d_1 \neq 0$. We may calculate that there are also $45$ numbers such that $d_1 + d_3 = d_2$, but there are $54 = 45 + 9$ numbers such that $d_2 + d_3 = d_1$. This would lead to $45 + 45 + 54 = 144$. However, some numbers are counted multiple times: those of the form $dd0$ and $d0d$. There are $18$ such numbers, so the solution is $144-18 = 126$. $\endgroup$ – TastyRomeo Jul 27 '16 at 12:42
  • $\begingroup$ @SteamyRoot You should post your comment as an answer. $\endgroup$ – N. F. Taussig Jul 27 '16 at 13:11

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