The integers from 200 down to 1 are written consecutively to form the large number N = 200199198197...131211...654321. Find the highest power of 3 which divides N.
Last edited by mr fantastic; Mar 21st 2011 at 06:26 PM. Reason: Re-titled.
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A quite modest 3^1
Can you please explain your reasoning behind this answer?
Probably one way to solve this problem is to keep in mind that 9 divides if and only if the sum of its digits is divisible by 9. Can you find a way to sum up the digits of this number to use this test?
Originally Posted by chris520 Can you please explain your reasoning behind this answer? Wrote program. Sum digits = 1902; divisible by 3, not by 9.
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