How many of the following octal (base 8) numbers are divisible by 1 4 1 0 (decimal)?
A. 3 4 6 0 3 4 3 6 1 0 7 5 3 6 2 5 2 3 6 5 0 2 3 5 6 4 2 0 3 3 5 4 3 5 3 5 0 1 1 3 5 3 4 5 4 4 2 8
B. 2 3 4 2 5 3 4 0 7 4 5 6 0 2 3 4 2 5 4 6 0 7 4 5 2 0 1 3 4 3 6 3 2 5 2 5 5 6 7 7 0 4 3 5 2 3 5 0 6 4 5 8
C. 2 3 5 7 6 7 3 4 2 3 5 0 2 3 5 6 3 4 5 2 3 4 0 5 2 3 4 0 2 5 3 0 5 2 3 5 0 0 5 3 4 5 2 0 0 0 5 3 5 0 0 0 8
D. 2 3 0 4 3 0 4 6 7 0 2 0 5 0 4 7 5 0 0 1 0 0 3 3 6 3 6 0 7 0 7 0 5 0 2 0 3 5 3 4 5 6 4 3 0 5 4 6 5 0 3 4 5 6 8
E. 3 5 7 4 5 6 0 4 5 6 3 5 4 2 3 5 0 4 6 5 4 6 6 5 7 0 4 3 5 5 4 3 5 0 6 5 7 3 4 5 0 4 5 5 0 6 7 8
F. 4 7 0 2 3 4 5 0 3 4 6 2 0 3 4 5 3 2 5 0 4 7 6 6 5 0 5 7 5 0 6 4 0 5 7 7 5 0 6 4 0 3 4 6 0 6 7 5 7 0 8
G. 5 4 7 0 4 5 2 5 4 6 0 4 6 7 0 6 7 5 6 0 6 3 5 5 3 4 6 0 4 0 5 3 2 0 1 0 1 4 2 3 4 0 1 5 2 3 0 5 2 1 5 4 8
H. 3 2 6 0 4 3 5 0 3 6 4 5 7 5 4 6 4 3 2 4 3 4 6 4 6 0 4 7 5 4 0 5 4 2 5 3 4 6 6 4 5 0 3 2 4 5 6 0 4 8
I. 1 4 6 7 0 5 4 5 7 0 7 0 7 0 4 5 6 7 7 0 0 6 4 5 6 7 0 4 5 0 6 4 5 6 0 4 2 1 0 2 3 4 1 2 3 5 1 5 0 3 2 8
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I'll leave the proof to others, but here is the trick to solve this problem. 14 = 2 * 7, so the number is divisible by 14 if it is divisible by 2 and by 7.
i. 2 divides 8, so an octal number is divisible by 2 if the last digit is even (like with base 10).
ii. In base 10, a number is divisible by 9 if the sum of the digits of the number is divisible by 9; likewise, in base 8, a number is divisible by 7 if the sum of the digits is divisible by 7. If you sum octally, you can keep adding the digits, and if the result is 0 or 7, the number is divisible by 7, otherwise not. E.g. the sum of E is 196 (decimal), that is 304 (octal), and 3 + 4 = 7.
The (decimal) digitsums are resp.: 159, 185, 155, 166, 196, 189, 172, 187 and 178. 196 and 189 are divisible by 7, but E is an odd number; only F is divisible by 2 and 7, and therefore by 14.