The six-digit number 7x1yyx is a multiple of 33 for non-zero digits x and y. Which of the following could be a possible value of (x + y)?
Show answer & explanation
Correct answer: A
Explanation
The correct answer is option 1 i.e. 5.
7x1yyx
It must be a multiple of 33.
Since 33 = 3 × 11, the number must be divisible by both 3 and 11.
A number is divisible by 3 if the sum of its digits is divisible by 3.
Digits of the number 7x1yyx are:
7, x, 1, y, y, x
⇒ Sum = 7 + x + 1 + y + y + x = 8+ 2x + 2y
So,
Divisible by 3 ⇒ 8 + 2x + 2y ≡ 0 mod 3
Divisibility by 11:
Rule: The Alternate sum of digits must be divisible by 11:
Number: 7 x 1 y y x
⇒ (7 − x + 1 − y + y − x) = 8 − 2x ⇒ 8 − 2x ≡ 0 mod 11
⇒ 2x ≡ 8 mod 11 ⇒ x ≡ 4 mod 11 ⇒ x = 4 (since x is a digit 1–9)
Number: 7 x 1 y y x ⇒ (7 − x + 1 − y + y − x) = 8 − 2x ⇒ 8 − 2x ≡ 0 mod 11 ⇒ 2x ≡ 8 mod 11 ⇒ x ≡ 4 mod 11 ⇒ x = 4 (since x is a digit 1–9)
Now put x = 4
Then check divisibility by 3 again:
Sum of digits = 8 + 2(4) + 2y = 8 + 8 + 2y = 16 + 2y
16 + 2y ≡ 0 mod 3 ⇒ 2y ≡ −16 ≡ 2 mod 3 ⇒ 2y ≡ 2 mod 3 ⇒ y ≡ 1 mod 3 ⇒ y = 1, 4, or 7
Try possible values of y:
Try x = 4, y = 1
Number = 741114
Check divisibility by 3: sum = 7 + 4 + 1 + 1 + 1 + 4 = 18
Check divisibility by 11: alt sum = 7 − 4 + 1 − 1 + 1 − 4 = 0
So 741114 is divisible by both 3 and 11 → divisible by 33
x = 4, y = 1 ⇒ x + y = 5