What is it?
What is the largest positive integer marven that can be formed from the digits 1, 2, 3, 4, 5, 6, each used exactly once, such that:
marven is divisible by 6,
marve is divisible by 5,
marv is divisible by 4,
mar is divisible by 3, and
ma is divisible by 2.
The answer is 321654.
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Since ma, marv and marven need to be divided by 2, 4 and 6, a, v and n need to be even numbers. So, m, r and e are odd numbers.
Since marve is divisible by 5, e needs to be 5. Thus e=5
So, for m and r, there are two possible values remaining: 1 and 3. Since m=3 would make bigger number than m=1, lets take m=3 and r=1.(If this is not possible, we will take m=1 and r=3)
If a=6, or a=4, mar is not divisible by 3. So, a=2.
Now, v and n has two possible values: 4 and 6. If v= 4, marv is not divisible by 4 (Just check the divisibility by 4 of last two digits). Thus, v=6 and n=4.
So, m=3, a=2, r=1, v=6, e=5 and n=4. So, the required number is 321654.