Find The Mod

Because $36|N$ , we must also have that $9|N$ , implying that the digit sum of $N$ must be a multiple of 9 (in base 10). All of $N$ 's digits are even, and so the digit sum must be a multiple of 18. Let's find the sum of all possible even digits.

$\quad0 + 2 + 4 + 6 + 8 = 20$

So the digit sum must be either 0 or 18, as it cannot be greater than 20. Notice that $20 - 2 = 18$ , and thus we can simply remove the digit 2. Removing anything else would either leave the sum unchanged (in the case of 0) or bring it below 20.

The largest integer with the digits $[0, 4, 6, 8]$ is 8640, and we can easily check to see that this number is truly a multiple of 36.

Thus, the answer is $8640 \equiv \boxed{640} \pmod{1000}$

If all digits are even and unique, the only possible digits are $2, 4, 6, 8, 0$ . But since $36$ is divisible by $3$ , the sum of the digits of any multiple of $36$ has to be divisible by $3$ .

Adding all the five possible digits, we get $2+4+6+8+0 = 20$ which is not divisible by $3$ . Clearly, not all five digits can be used. Note that we want the greatest multiple. We now search for four digit multiples. But which of the five digits do we eliminate?

1. If we eliminate 0, the sum of the remaining digits is 20 (indivisible by 3)

2. If we eliminate 2, the sum of the remaining digits is $18$ (divisible by 3)

3. If we eliminate 4, the sum of the remaining digits is 16 (indivisible by 3)

4. If we eliminate 6, the sum of the remaining digits is 16 (indivisible by 3)

5. If we eliminate 8, the sum of the remaining digits is $12$ (divisible by 3)

Hence, the greatest integer multiple has to compose of either the digits $0468$ or $0246$ . Since we want the greatest integer multiple, let us consider the largest possible four digit value between these sets: $8640$ .

$8640 \div 36 = 240 \Longrightarrow$ $8640$ is a multiple of 36, and thus the multiple we want.

$8640 \mod{1000} = \boxed{640}$

The digits in the number are from the set { $0,2,4,6,8$ }. This means that the number has at most five digits. Since the number must be a multiple of $36$ , we know that is must be divisible by $4$ and $9$ .

For a number to be divisible by $9$ , the sum of its digits must be a multiple of $9$ . We want to maximize the number, so we try using all the digits. However, the sum of the digits would be $0+2+4+6+8=20$ , leading to the number not being divisible by $9$ . The closest multiple of $9$ less than $20$ is $18$ . It follows that the digit $2$ will not be in the number.

Now we need to make the number divisible by $4$ while using the digits $0,4,6,8$ . With a little experimentation, we find the largest number to be $8640$ . The remainder when the number is divided by $1000$ is $\boxed{640}$ .

When all of the digits are even and differents, the digits could be 2,4,6,8,0 but 2+4+6+8+0=20 => It's imposible get a number with 5 digits, we have that 18 is the closer number that 20 => we have the digits 4,6,8,0 and the numer 8640 is the biggest with this digits and is multiple of 36 => the answer is 640.

No matter the order of the digits of the number $86420$ , that number will not be divisible by $36$ , because the sum of it's digits isn't divisible by $9$ . So we have $5$ combinations of $4$ digits that we can choose from the digits $86420$ . The only combination divisible by $9$ is $8640$ . It is also divisible by $4$ , because $40$ is divisible by $4$ . Hence, $N = 8640 \equiv \boxed {640} \pmod {1000}$ .

If $N$ is a multiple of 36, it must be divisible by $9$ and $4$ .
The even digits possible are $0, 2, 4, 6, 8$ which sums to 20. We omit $2$ then. Now we are left to permute the digits $0, 4,6,8$ such that it is a multiple of $4$ . To have the greatest number, we can have $40$ as the last two digits, since it satisfies the divisibility rule for 4. Thus we have the number $8640 \equiv \boxed{640} \pmod {1000}$ .

Even numbers are 2,4,6,8,0; so the number \­[ \N \­] will have 5 numbers as most.

\­[ \N \­] is multiple of 36, so it's also multiple of 3. We know that the digit sum of multiples of three are also multiples of 3. If we do: 8+6+4+2+0; we get 20, that is not a multiple of 3.

So now, \­[ \N \­] will have 4 numbers maximum. If we remove number two: 8+6+4+0=18, that is a multiple of three.

If we order these four numbers we have 8640. \­[ \frac{8640}{36} = 240 \­].

The answer, then, taking the last three numbers, is \­[ \boxed{640} \­]

As 36 divides N, we have 5 options for its unit digit.: 0, 2, 4, 6 and 8. But as it should not have any digit repeated, the maximum digits it may have is just 5. For 5 digits, its sum will be 20 which is clearly not divisible by 9 which is required condition for a number to be divisible by 9. In 4 digits no, we see that dropping 2 makes our number divisible by 9 and hence digits of N are 8, 6, 4 and 0. So 8640 is required number and 640 is the answer...

8640 will be the biggest no.,Last three digits will be the remaninder..

$N$ is a permutation of one or more of the following digits $0,2,4,6,8$ . So numbers of the form $\{86420, 68420 \dots \}$ .

We can write $36$ as $2^2 \times 3^2$ . All even numbers are divisible by $2$ , so now we have to find the largest even number having all even (different) digits. We want this number to be as large as possible.

How about $86420$ ? This isn't divisible by $3$ as $8 + 6 + 4 + 2 + 0 = 20$ isn't divisible by $3$ .

Now we want the number to be as large as possible, but excluding one digit. Thus, we get $8640$ . Now, $8640$ is divisible by both $2$ and $3$ . We have $2$ twos and $2$ threes. Dividing $\dfrac{8640}{2\times 3} = 1440$ . Now, $1440$ is divisible by $2$ and $3$ too. Thus, we've got our number which is $8640$ .

Now, $8640 \mod{1000} = \boxed{640}$ .

If we include all the even digits for the greatest integer multiple, we find that it is impossible for it to be divisible by 3, therefore by 36 as well. The next logical try would be 8640, which happens to be divisible by 36. Thus N = 8640 mod 1000 = 640

N should be divisible by 9 as $36=9*4$ so digits will be 4,6,8,0 because $4+6+8+0=18$ last 2 digits should be divisible by 4 and $8640$ is the maximum number can be made under that conditions.

unique even elements 0,2,4,6,8<br> if it divisible 9 sum is 9<br> 8640 maxnumber<br> so answer is 640<br>

since all of the digits are even ,therefore the possible digits are 0,2,4,6,8 and for the number to be divisible by 36 it ought to be divisible by 6 and in turn by 2&3 .the number having even digits is obviously divisible by 2 and thus to be divisible by 3 sum should be divisible by 3 and be max. thus possible sum is 18 and the number would be 8640 thus remainder 640

The greatest integer with even digits that are not repeated is 86420, which isn't divisible by 3, hence not divisible by 36. The next greatest integer that satisfies the above criteria in addition to being divisible by 3 is 8640, which turns out to be divisible by 36. So now we have a multiple of 36 which could be a possible answer. When you divide this number by 1000, the quotient is 8 with 640 as the remainder.

$36=9\times4$ ,So if a number will be divisible by 36 if and only if the number is divisible by both $9$ and $4$ .

A number will be divisible by $9$ if and only if the digit sum of the number is divisible by $9$ .

Without repeatation, from the even digits, the sum will be divisible by $9$ when we take $8,6,4,0$

$8+6+4+0=18,\frac{18}{9}=2$

The maximum number formed by these digits is $8640$ , which is also divisible by $4$ .

So $N=8640$ ,and the remainder is $\boxed{640}$

You can see that $36 = 2^{2}*3^{2}$ , so 9 and 4 will be the divisor of $N$ .

Since the digits are all even, consider the $2,4,6,8,0$ .

• $N$ has a factor of $9$ $<=>$ the digit sum has a factor of $9$ .

So there's only 1 case that satisfy this statement. $(4,6,8,0)$ because $9|4+6+8+0=18$

Therefore: the maximum of $N$ is $8640$ . Answer $\boxed{640}$ ~~~

For the number to be divisible by $36$ it should be divisible by $9$ and $4$ . The largest possible number with distinct even digits is $86420$ . This is however not divisible by $9$ as according to the test of divisibility of $9$ the sum of the digits should be a multiple of $9$ . The sum of the digits of $86420$ is $8+6+4+2+0=20$ . We want to remove as few numbers as possible to make the sum divisible by $9$ . This is easily done by removing $2$ . This number $8640$ also happens to be divisible by $4$ as according to the divisibility test of $4$ the last two digits should be divisible by $4$ . Hence our answer is $\boxed{640}$