If the seven-digit number 5x634y2 is divisible by 88 such that x

=y, then what is the value of (2y−3x)?

For a number to be divisible by 88, it must be divisible by both 8 and 11.
Step 1: Divisibility by 8
The last three digits must be divisible by 8. The last three digits are 4y2.
Let's check values of y (0≤y≤9):
If y=3, 432÷8=54 (Divisible)
If y=7, 472÷8=59 (Divisible)
So, y can be 3 or 7.

Step 2: Divisibility by 11
The difference between the sum of digits at odd positions and even positions must be 0 or a multiple of 11.

Sum of odd positions (from right)=2+4+6+5=17
Sum of even positions (from right)=y+3+x=x+y+3
Difference=17−(x+y+3)=14−(x+y)
For divisibility by 11, 14−(x+y) can be 0 or 11.

Case 1: If y=3

14−(x+3)=11⟹11−x=11⟹x=0
14−(x+3)=0⟹11−x=0⟹x=11(not possible as x is a single digit)
So, if y=3, then x=0. Here x

=y condition holds.
Let's evaluate (2y−3x):

2(3)−3(0)=6
Case 2: If y=7

14−(x+7)=11⟹7−x=11⟹x=−4(not possible)
14−(x+7)=0⟹7−x=0⟹x=7
But the problem states x

=y, so x=7,y=7 is rejected.

Thus, x=0 and y=3.

Value of (2y−3x)=2(3)−3(0)=6