find the least values of x and y so that 7x 342y is divisoble by 88 

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 )