# Thread: n = 13*(sum of digits)

1. ## n = 13*(sum of digits)

The number of positive integers which are 13 times the sum of their digits is

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

This is some competition problem, AMC(2000). I know I should start by writing out the decimal expansions but I don't know how many digits the number has or what to do next.

2. Originally Posted by DivideBy0
The number of positive integers which are 13 times the sum of their digits is

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

This is some competition problem, AMC(2000). I know I should start by writing out the decimal expansions but I don't know how many digits the number has or what to do next.
Well it obviously isn't true for 1 digit. Try 2:
"ab" = 10a + b

So we want x, y such that
$10a + b = 13(a + b)$

So
$b = -\frac{a}{4}$
which is ridiculous.

So try 3 digits, abc:
$100a + 10b + c = 13(a + b + c)$

So
$c = {(100 - 13)a - (13 - 10)b}{12} = \frac{29a - b}{4}$

This, at least, might be possible. We need 29a - b to be divisible by 4 with a and b in the range from 0 - 9. By trial and error I immediately run into (a, b) = (1, 1):
$c = \frac{29 - 1}{4} = 7$

So 711 fits the bill.

See if you can find others.

What can you say about a 4 digit number?
$1000a + 100b + 10c + d = 13(a + b + c + d)$

Solve this for d and see if you can find any a, b, c for it and why/why not.

-Dan

3. Ok, that means...

If N is a 4-digit number then

$1000a+100b+10c+d=13(a+b+c+d)$

$\Rightarrow d = \frac{329a-29b-c}{4}$

Even if you have a = 1, b = 0, c = 9... d = 80. Any other values of $a\neq 0$, b, c only give larger values of d. I'm guessing this would only be exacerbated if N had > 4 digits.

So N must have 3 digits

Hohoho, I think there's a good way of doing this.

As you said, $c=\frac{29a-b}{4}$

Now a can only be 1, if it is any greater than we get the same trouble we had with 4 digit numbers, c will be too great.

Then $29(1)-b \equiv 0$ (mod 4), so $29 \equiv b$ (mod 4). That is, $b\text{mod}4=1$. So $b=1,5,9$

Yes! So the solutions are 195, 117, 156. (D)

4. Originally Posted by DivideBy0
Ok, that means...

If N is a 4-digit number then

$1000a+100b+10c+d=13(a+b+c+d)$

$\Rightarrow d = \frac{329a-29b-c}{4}$

Even if you have a = 1, b = 0, c = 9... d = 80. Any other values of $a\neq 0$, b, c only give larger values of d. I'm guessing this would only be exacerbated if N had > 4 digits.

So N must have 3 digits.

Hohoho, I think there's a good way of doing this.

As you said, $c=\frac{29a-b}{4}$

Then $29a-b \equiv 0$ (mod 4), so $29a \equiv b$ (mod 4). That is, $b\text{mod}4=1$. So $b=1,5,9$

3 Solutions (D)!
Bingo!

-Dan

5. Originally Posted by DivideBy0
This is some competition problem, AMC(2000). I know I should start by writing out the decimal expansions but I don't know how many digits the number has or what to do next.
Which AMC? I presume this is AMC-10. You are in the 10th grade right?

6. Yep, I'm in grade 10 and that's the Australian MC