# Using the digits 4, 5, 6, 7, 8, each once, construct a 5-digit number which is divisible by 264.

First, factorize 264.

Factors of 264 = 2 × 2 × 2 × 3 × 11

Or factors of 264 = 23 × 3 × 11

So, if a number is to be divisible by 264, then the number must be divisible by 2, 3 and 11 necessarily.

For divisibility by 2,

The number should contain an even number in the unit digit.

So, using the given numbers 4, 5, 6, 7, 8, we can say that the number so formed will have either 4 or 6 or 8 in the end.

For divisibility by 3,

The sum of the digits of the number must be divisible by 3.

4 + 5 + 6 + 7 + 8 = 30

30 is divisible by 3.

For divisibility by 11,

The difference of the sum of digits at odd places and sum of the digits at even places must be equal to a multiple of 11.

Sum of all numbers = 4 + 5 + 6 + 7 + 8

Sum of all numbers = 30

Let us try to form a combination whose sum of digits at even places and sum of digits at odd places is equal to 15.

Such combination of numbers are {8, 7} and {4, 5, 6}.

Now, form a 5-digit number using these two groups, provided the unit digit be either 4, 6 or 8.

For unit digit of 4, we have 68574, 57684, 67584, 58674.

For unit digit of 6, we have 48576, 58476, 57486, 47586.

And the number 8 can’t be used in the unit place.

So, we have 5-digit numbers namely, 68574, 57684, 67584, 58674, 48576, 58476, 57486 and 47586.

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