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Let $N$ be the greatest integer multiple of $36$ all of whose digits are even and no two of whose digits are the same. Find the remainder when $N$ is divided by $1000$.

$$N = \overline{abcd....} = 36x$$

Even numbers: $0, 2, 4, 6, 8$

A maximum number that can be made is of $5$ digits.

$36 \cdot 12 = 423$, $36 \cdot 24 = 864$, $36 \cdot 13 = 468$, $36 \cdot 13 = 504$

Since $36 = 4 \cdot 9$, a number is divisible by $9$ if the sum of the digits are divisible by $9$.A number is divisible by $4$ if the last two digits are divisible by $4$.

Lets list out multiples of $9$:

$9, 18, 27, 36, 45, 54,...$

Sum{max}: $8 + 6 + 4 + 2 = 20$ Sum{min}: $0, 2$.

$8 + 6 + 4 = 18 = 2(9)$.

I can't find a number since, $4$ doesnt divide $18$?

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It is the number which has to be a multiple of $4$, not the sum of its digits. For example, the number $468$ (note that this is not the solution to the full problem) has a digital sum of $18$ (and thus is divisible by $9$) and ends in $68 = 4\cdot 17$ (and thus is divisible by $4$).

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You almost solved it. Since $8+6+4+2+0=20$ is not divisible by 9, this number must only have 4 digits. And the greatest multiple of 9 with 4 distinct even digits is 8640, which is also divisible by 4. So the answer is 640.

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