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How many five-digit integers exist with the following conditions,

$1.$ be divisible by $125$

$2.$ Sum of their digits be divisible by $4$.

$1) 176 \qquad\qquad 2)178 \qquad\qquad 3)180 \qquad\qquad 4)198\qquad\qquad 5)200$

To solve this problem I noted that the last three digits of these numbers should be either one of the following, $000, 125, 250, 375, 500, 625, 750, 875$. Now if I divide the sum of each of these three digits by $4$, the remainder will be, $0, 0, 3,3,1,1,0,0$ respectively. Now for four cases, sum of the first two digits should be divisible by four, and for two cases it should be in the form $4k+1$ and finally for the last two cases it should be $4k'-1$.

Now after counting all these cases, which was relatively time-consuming, I got $178$ as the answer.

I'm wondering if it is possible to solve this problem with alternative approaches. Since sum of the digits of a number is related to the remainder of dividing a number by $9$ I think this idea would be helpful, but not sure how to implement it.

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  • $\begingroup$ This is the way. Sum of the digits works for divisibility by $3$ and $9$ since $10^n-1$ is divisible by $9$ and one can write every number $d_nd_{n-1}\dots d_1d_0$ as $\sum_{i=0}^nd_i\cdot10^i$. So that won't be of any help for divisibility by $4$. $\endgroup$ Jan 7 at 1:22

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There are $720$ 5-digit numbers that are divisible by 125.

We would expect the digital sums to be approximately uniformly distributed, so close to 1/4 should have a digital sum that is divisible by 4.

The answer should be close to 180. If you wanted to guess and move on, you could take a stab.

But, what is it exactly?

As noted, the last 3 digits must be $\in \{000,125,250,375,500,625,750,875\}$ The associated remainders with these suffixes are $\{0, 0, 3, 3, 1, 1, 0, 0\}$

What are the two-digit prefixes?

With the prefixes $10 - 89$ we have prefixes such that the digitals sums will be uniformly distributed. i.e. there are 20 prefixes associated with each remainder.

Or, there are $160$ numbers in the range $10,000 - 89,000$ divisible by 125 with digital sums divisible by 4.

We must count the numbers for the prefixes in the range $90 - 99$ independently.

$\{90,94,98\}$ have remainder 1
$\{91,95,99\}$ have remainder 2
$\{92,96\}$ have remainder 3
$\{93, 97\}$ have remainder 0

Count the number that is required for each remainder.

$2+2+2+2+3+3+2+2 = 18$

$178$

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