Question

2Y2X56 IS DIVISIBLE BY 11 AND 8. HOW MANY VALUES CAN Y TAKE?

Answer 1

Answer:

2 values which are either X = 0 and Y= 3 or X = 8 and Y =  6

Step-by-step explanation:

Hi,

Given 2Y2X56 is divisible by both 8 and 11

Divisibility Rule by 8

=> For a number to be divisible by 8, its last 3 digits should be divisible 8 ,

=> X should be either 0 or 8.

Now. for a number to be divisible by 11, difference of sum of its alternative sums of digits should be divisible by '0'

=>(2 + 2 + 5) - (Y + X + 6) should be divisible by 11

=> X + Y -3 should be divisible by 11

Now, if X = 0

Y - 3 should be divisible by 11

This happens only if Y = 3.

Similarly, if X = 8,

then Y + 5 should be divisible by 11,

This happens only when Y =  6.

Hence, the possible values of Y are 2, which are

X = 0 and Y= 3 or X = 8 and Y =  6.

Hope, it helped !

Answer 2

Answer:

4 cases are possible.

X=2, Y=1

X=0, Y=3

X=8, Y=6

X=6, Y=8

Step-by-step explanation: