Re: another GRE math exercise problem

"john" <johnboy98105@xxxxxxxxx> wrote in message
news:1166699026.052615.34260@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
thanks everyone. I have to say adding up the digits is a new tirck for
me. I've never seen that...

MY guess is that it works for other numbers? If the digits add up to
say a number divisible by some other smaller integer, does that mean
the larger number is divisible by the smaller digit sum?

Sorry for a strange question.

It isn't that simple. It has to do with:

a)The value of the the prospective divisor (3, for the purposes of example),
versus the radix of the representation (10 for us humans).
b)The prime factorization of the prospective divisor vs. the radix, and LCM,
GCD, etc.

For example, because 10 mod 9 is 1, any multiple of 3 will have digits that
sum to a multiple of 3. Essentially what happens (considering only the
2-digit case) is that when you roll over from 9 to 12, the ten's digit is
incremented by the value (1) that the modulo value of the one's digit loses
(1).

You'd find this in any case where LCD(divisor, radix) = 1 that modulo
relationship holds. For example, assume we are counting in base-5 and we
are interested in divisibility by 4.

Let's choose a good example: 1030_base10 is 13110_base5. We know of course
that 1030_base10 is a multiple of 2. Adding the digits of 13110_base5
together in base5 gives 11_base5, and adding again gives 2_base5, which is
the expected result.

But in the general case, the sum of digits won't indicate divisibility ...
10 and 3^3 is a very special case, as is 5 and 2^2.

Dave.



.

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