Julia Thompson wrote: > > In base N, to check to see if a number is divisible by N-1, just add the > digits, and if their sum is divisible by N-1, the number itself is. So > in base 10, if the sum of the digits of a number add up to 9 or 18 or > 27, etc., the number is divisible by 9. > > If N-1 is a square, a similar divisibility test will work on sqrt(N-1). > So if the sum of digits of a number in base 10 is divisible by 3, the > number itself is divisible by 3. > No, the correct form of the test is: if N-1 is not a prime number, then the same test works for the _divisors_ of N-1.
And there's also a simple test for N+1 and (N+1)'s divisors: if the sum of the even-ordered digits and the sum of the odd-ordered digits differ by a multiple of the divisor under test, then the number is divisible by it. For example, 8074 is divisible by 11 in base 10. > Base 12 has easier divisibility tests for more numbers, though. > Yes, but we gain useless numbers, like 13, and lose useful numbers, like 5. Alberto Monteiro _______________________________________________ http://www.mccmedia.com/mailman/listinfo/brin-l
