To be divisible by 4, last 2 digits must be div by 4. So y can be 0, 2, 4, 6 or 8.

Now to be divisible by 11, |(the sum of the digits in even places) – (the sum of the digits in odd places)| should be divisible by 11 (or should be 0)

Sum in odd places is 18 + x and in odd is 13 + y, trying out the various values of y we find that when y = 0, x = 6 to satisfy the condition, similarly
when y = 2 x = 8,
when y = 4, no value of x exists
when y = 6, x = 1 and
when y = 8, x = 3.

This is cool !

Off topic: What is the least value of x if the nine digit number 23x4567y4 is divisible by 44 ?

Your help would be really beneficial. Thank you.

23x4567y4 is divisible by 44 i.e. 11 and 4.

To be divisible by 4, last 2 digits must be div by 4. So y can be 0, 2, 4, 6 or 8.

Now to be divisible by 11, |(the sum of the digits in even places) – (the sum of the digits in odd places)| should be divisible by 11 (or should be 0)

Sum in odd places is 18 + x and in odd is 13 + y, trying out the various values of y we find that when y = 0, x = 6 to satisfy the condition, similarly

when y = 2 x = 8,

when y = 4, no value of x exists

when y = 6, x = 1 and

when y = 8, x = 3.

So minimum x is 1 I guess.

regards

J

Thanks a lot. You guys rock !

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J