What is the largest 3-digit number which has all 3 digits different and is equal to 37 times the sum of it's digits?

-Question 28, Junior Division, AMC 2016

I found the solution here, page 51, however I don't understand it from the very beginning. How does $100a+10b+c$ equal $37a+37b+37c$? Can somebody explain to me what's happened in this solution step by step please? Or can somebody give an alternate solution, suitable for Year 7 and 8? Thanks!

P.S. If anybody knows a more accurate tag for this question, please feel free to edit my question or comment so I can edit it if you can't.

  • $\begingroup$ If $a,b,c$ are the digits of the number, then the number is $100a+10b+c$, the sum of the number's digits is $a+b+c$, and $37$ times the sum of its digits is $37(a+b+c)=37a+37b+37c$. $\endgroup$ – arctic tern Jul 26 '17 at 6:13
  • $\begingroup$ If the three digits are in order (from left to right) $a$, $b$ and $c$, $37a+37b+37c$ is 37 times the sum of the digits, and $100a+10b+c$ is the number. If that's where your problems start, my guess is that AMC is not for you. $\endgroup$ – Henrik supports the community Jul 26 '17 at 6:14
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    $\begingroup$ Um... That link is to a file on your computer. We can't actually read it. $\endgroup$ – fleablood Jul 26 '17 at 7:52
  • $\begingroup$ @fleablood - Sorry ¯_(ツ)_/¯. Problem fixed! $\endgroup$ – bio Jul 26 '17 at 9:36

You have $\overline{abc}=100a +10b+ c=37 (a+b+c)$ or

$$63a-27b-36c=0 \ \ \Longrightarrow 7a=3b+4c$$

Now $a \le \frac{3b+4c}{7}\le \frac{3\times 8+4 \times 9}{7}<9$.

Let $a=8$. Then $3b+4c=56$. We have $3b=4(14-c)$. Thus $b$ is a multiple of $4$. This forces that $b=4$ and gives $c=11$, which is impossible!

Let $a=7$. Then $3b+4c=49$ and $b=\frac{49-4c}{3}=16+\frac{1-4c}{3}$. This gives $c=4$ and leads to $b=11$, which is impossible again.

Let $a=6$. Then $3b+4c=42$ and $4c=3(14-b)$. This gives $c=6, 9$ since $c$ must be a multiple of $3$. Then possible pairs for $(b, c)$ are $(6, 6), (2, 9)$.

Thus largest is $629$.

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"What is the largest 3-digit number...."

Let $N = abc$ have three digits. $N = 100a + 10b + c$. That's what the writing numbers of digits mean. $593 = 500 + 90 +3$ and $abc = 100a + 10b + c$.

" which has all 3 digits different"

So $a \ne b; a\ne c; b \ne c$.

" and is equal to 37 times the sum of it's digits?"

So $N = 37*(a+b+c) = 37a + 37b + 37c$

And $N = 100a + 10b + c$.

So $100a + 10b + c = 37a + 37b + 37c$

So $(100a - 37a)= (37 - 10)b + (37-1)c$

So $63a = 27b + 36c$

So $\frac {63}9 a = \frac {27}9 b + \frac {36}9 c$

So $7a = 3b + 4c$.

Obviously $a = 1, b= 1, c=1$ is a solution but 1) It's probably not the largest and 2) The digits aren't all different.

$b \ne c$ so $c = b \pm k$ for some $k \ne 0$.

So $7a = 3b + 4c = 3b + 4(b \pm k) = 7b \pm 4k$

So $a = b \pm \frac 47k$ so $7$ must divide $k$. But $b$ and $c$ are single digits and $c = b \pm k$ so $k = 7$ and $a = b \pm 4$

The possible answers are $b =0; c = 0+7=7; a = 0+4=4$

$b=1; c = 1+7= 8; a = 1+4 =5$

$b=2; c = 2+7 =9; a= 2+4 =6$

$b = 7; c = 7-7=0 ; a=7-4 = 3$

$b = 8; c = 8-7; a = 8-4 = 4$

$b = 9; c = 9 -7 =2; a = 9-4=5$.

Of those $a=6;b=2; c= 9$ and $N = 629 = 37*17 = 37(6+2+9)$ is the largest such answer.

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Since $3\cdot 37=111$ and all three-digit multiples of that have three equal digits, our number better not be a multiple of $3$.

The digit sum is at most $9+9+9=27$, so (cf. first paragraph) out number is at most $26\cdot 37=962$. However, $9+6+2\ne 26$. Any number $<962$ has digit sum $\le 9+5+9=23$, so our number is at most $23\cdot 37=851$. Again, $8+5+1\ne 23$, so we must go lower again. Actually, this is just a finite problem - why not simply check the few multiples of $37$ up to that limit?

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The notation in which all our numbers are written is called the place value notation. You may have had exercises in class 5 and 6, maybe, concerning decimal expansions of the a number. For example, $678 = 6 * 100 + 7 * 10 + 8 * 1$, and $1089 = 1 * 1000 + 8 * 100 + 9 * 1$.

Therefore, using elementary algebra, we can deduce that given a three digit number represented by $abc$ is the number $100a+10b+c$. Here, I say represented by because in ordinary algebra $abc$ is the product of the three quantities a,b, and c, but here we are treating it as the representation of a three digit number e.g. $678$.

From the representation $abc$, we deduce that $a,b,c$ stand for the digits of the number. For example, when we write $678$, then the digits of the number are $6,7,8$.

Now, we are in a position to actually understand the problem at least : it says which is the largest three digit number is $37$ times the sum of it's digits?

Let this number be $abc$. Note that this is a representation. The actual number, of course, is $100a+10b+c$, and this is equal to $37$ times the sum of digits which are $a,b,c$, hence giving the equation $100a+10b+c = 37a+37b+37c$.

When it comes to solving this question, you should realize that our three digit number is at least a multiple of $37$, the quotient being $a+b+c$. So it's enough to check multiples of $37$. Furthermore, since we have to find the largest such number, we may as well start from the largest three digit number possible, which is $999$. This in fact satisfies all the requirements, but doesn't have distinct digits.

So we come further down by subtracting $37$. The next number is $962$, doesn't work. The next is $925$, doesn't work. The next is $888$, doesn't have distinct digits, ... (once you understand why we are doing all this, I can tell you ways of reducing the number of cases to check in this step, because it's time intensive).

Work your way down, until you find the right answer, which I think is $37*17 = 629$. The sum of the digits is $17$, indeed, and the digits are distinct.

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let $$100a +10b+ c=37 (a+b+c)$$

on solving you get $$63a-27b-36c=0$$ consider the largest possible case that is let $a=b=9$ then on substituting the values in the equation you get $c=9$ ...bingo!!

i don't think there is any other specific way. hope it helps.

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  • 2
    $\begingroup$ note that all $3$ digits have to be different. $\endgroup$ – Siong Thye Goh Jul 26 '17 at 7:15
  • $\begingroup$ oh sorry had not seen it $\endgroup$ – AkashdeepKar Jul 26 '17 at 7:25

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