Question

If number 6y81x is divisible by 55,what are possible values of x and y? (No absurd answer please ,else will be reported and deleted)

Answer 1

*Answer:

For 5x423y to be divisible by 88, it must be divisible by 11 and 8.To be divisible by 8, 23y should be divisible by 8.To be divisible by 11, (5 + 4 + 3) - (x + y + 2) is 0 or a multiple of 11.12 - 2 - (x + y) = 0 or multiple of 11.10 - (x + y) cannot be a multiple of 11 , hence should be 0.This gives us the possible values of x and y as (9,1),(8,2),(7,3),(6,4),(5,5) and vice versa.We can also say that y can take a value of only 2, for 23y to be divisible by 8.The least value of x and y so that the number 5x423y is divisible by 88 is 8, 2.

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Answer 2

Answer:

ANSWER

For 5x423y to be divisible by 88, it must be divisible by 11 and 8.

To be divisible by 8, 23y should be divisible by 8.

To be divisible by 11, (5 + 4 + 3) - (x + y + 2) is 0 or a multiple of 11.

12 - 2 - (x + y) = 0 or multiple of 11.

10 - (x + y) cannot be a multiple of 11 , hence should be 0.

This gives us the possible values of x and y as (9,1),(8,2),(7,3),(6,4),(5,5) and vice versa.

We can also say that y can take a value of only 2, for 23y to be divisible by 8.

The least value of x and y so that the number 5x423y is divisible by 88 is 8, 2.