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#1 2006-03-15 01:43:44



Largest power of "K" problem...yikes

What is the largest power k such as 3^k divides easily into 40? With out anyremainders?
a. 1
b. 4
c. 13
d. 17
e. none of these

I know a can't be right as 40 is to even of a number to be that only.
but do I use the k like tha other problem with 1/2^3, 1/2^4...and so on? Or is there another one???

#2 2006-03-15 01:48:01

Full Member


Re: Largest power of "K" problem...yikes

I think you need to replace k with those numbers from a, b, c and d.
3^1 = 3
3^4 = 81
3^13 = 1594323
3^17 = 129140163

Since none of these divide into 40 without a remainder, I'd say the answer is e.

Aloha Nui means Goodbye.

#3 2006-03-15 02:55:16



Re: Largest power of "K" problem...yikes

rickyoswalidow is absolutely right!
3^k is a power of 3 and the factors of this number are 1, 3 and the higher powers of 3 up to 3^(k-1). For a number to be divisible by 40, the number should have three 2s and 5 as prime factors. Since 3^k does not have these prime factors,it is not divisble by 40 for any value of k!

Character is who you are when no one is looking.

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