Question
If the 5-digit number 693XY is divisible by 3, 7, and
11, then what is the value of X + 2Y?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
More ESIC Part B Questions
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17, 18, 22, 31, 47, ___
