If the 5-digit number 693XY is divisible by 3, 7, and 11,

Solution

ATQ;
The 5-digit number 693XY is divisible by 3, 7, and 11.

Divisibility rule of 3: The sum of digits must be a multiple of 3.

6+9+3+X+Y=18+X+Y

So, X + Y must be one of {0, 3, 6, 9, 12, 15, 18}.

Divisibility rule of 11: The difference between the sum of alternate digits must be divisible by 11.

(6+3+Y)−(9+X)=0

9+Y−9−X=0

Y−X=0

X=Y

From X + Y = 6, and X = Y, we get:

2X=6⇒X=3,Y=3

Checking divisibility by 7:

The number 69333 is divisible by 7.

Finding X + 2Y:3+2(3)=3+6=9