# math question?

all the six-digit numbers that can be formed using the digits 2,3,5,6,7,9 are written. If each digit can be used only once, what is the largest of these numbers that are divisible by 11

Relevance
• Start working backwards from the largest permutation and seeing if the remainder is 0:

permutation, remainder

976532, 7

976523, 9

976352, 3

976325, 9

976253, 3

976235, 7

975632, 9

975623, 0 <--

975623

Alternatively, use the rule of divisibility by 11:

An integer number is divisible by 11 if and only if the alternate sum of its digits is divisible by 11.

In other words, add up the 1st, 3rd and 5th digits then subtract the 2nd, 4th and 6th digits. The result must be divisible by 11.

We want the first digit to be as large as possible, so it should be 9.

We want the second digit to also be as large as possible, so it should be 7.

9 7 _ _ _ _

The total digit sum is 32, so the only way it will work is if the sum of the 1st, 3rd and 5th digits is 16 and likewise for the 2nd, 4th and 6th.

The 3rd and 5th digits must therefore add to be 7 and that only works if we use 5 and 2. We want them in that order with the larger digit first:

9 7 5 _ 2 _

And the remaining digits must be 6 and 3, again with the larger digit first.

9 7 5 6 2 3

Double-check:(9 + 5 + 2) - (7 + 6 + 3)

= 0 (divisible by 11).

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