In the following question, a number is coded as a pair of digits, where:
- The first digit is the remainder when the number is divided by 7.
- The second digit is the remainder when the number is divided by 11.
That is, if a number N yields a remainder 'a' when divided by 7 and a remainder 'b' when divided by 11, the code is (a, b).
Example:
If N = 53: 53 ÷ 7 = 7 with a remainder of 4;
53 ÷ 11 = 4 with a remainder of 9; thus, the code is (4, 9).
What is the smallest number with the code (4, 4) in this coding system?
Show answer & explanation
Correct answer: C
Explanation
N ÷ 7 has a remainder of 4
N ÷ 11 has a remainder of 4
That is:
N ≡ 4(mod 7)
N ≡ 4(mod 11)
This means:
N − 4 must be divisible by both 7 and 11.
LCM(7, 11) = 77
So:
N=77k+4
For the smallest number k = 1:
N=77+4=81
Check:
81 ÷ 7 = 11 remainder 4
81 ÷ 11 = 7 remainder 4
Correct answer: (3) 81