Question

Find the least value of 'X' and Y' so that 7X 342Y is divisible by 88.
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Answer 1

Answer:

To find least value of x and y of 7x342y is divisible by 88

If any number divisible by 88 , that number must be divisible by 8 and 11 both .

We know from divisibility rules of 8 if last three digits of any number is divisible by 8 then whole number also divisible by 8 .

So,

42y is divisible by 8

Only at y = 4 ( 424 ) is divisible by 8

So,

y = 4 Now our number : 7x3424

We know divisibility rule for 11 : If the number of digits is even, add the first and subtract the last digit from the rest.

Here we get even digits in our number ( 7x3424 ) , So

= x342 + 7 - 4

= x342 + 3

= x345 Again apply same rule and get

= 34 + x - 5

= 29 + x

So,

( 29 + x ) is divisible for least number 4 , So ( 29 + 4 = 33 that is divisible by 11 )

So,

Least value of x = 4 and least value of y = 4 ( Ans )

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