If a 10-digit number M30348462N is divisible by both 8 and 11, then what is the value of \(M^2+N^2-18\)?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 7
Two divisibility rules fix the two unknown digits. A number is divisible by 8 when the number formed by its last three digits is divisible by 8, and by 11 when the difference between the sums of its alternate digits is 0 or a multiple of 11.
Finding N (divisibility by 8):
The last three digits are 6, 2, N; of 620 to 629, only 624 is a multiple of 8 (\(8\times78\))
So N = 4
Finding M (divisibility by 11; the digits are M, 3, 0, 3, 4, 8, 4, 6, 2, 4):
Digits in the odd places: \(M+0+4+4+2=M+10\)
Digits in the even places: \(3+3+8+6+4=24\)
The difference \(24-(M+10)=14-M\) must be 0 or a multiple of 11
M is a single digit (0 to 9), so \(14-M\) lies between 5 and 14; the only multiple of 11 there is 11
\(14-M=11\), so M = 3
The value asked for: \(M^2+N^2-18=9+16-18=7\)