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by sukhman » Sat Oct 12, 2013 6:09 am
Find the maximum value of n such that 570*60*30*90*100*700*343*720*81 is perfectly divisible by 30^n
A. 12 B.11 C.14 D. 13
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by theCodeToGMAT » Sat Oct 12, 2013 7:34 am
Is the Answer [spoiler]{B}[/spoiler]

Steps:

30^n = 3^n * 10^n --> we need to find power of 3.. we can find that by factorizing each number

570 = 3 x 190
60 = 3 x 20
30 = 3 x 10
90 = 9 x 10
100 = NO 3
700 = NO 3
343 = NO 3
720 = 9 x 80
81 = 9 x 9
So, 1+1+1+2+2+4 ==> 11
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by sukhman » Sat Oct 12, 2013 8:13 am
can you plz make it clear why is not divided by 5 since 30=2*3*5 Though your answer is correct
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by theCodeToGMAT » Sat Oct 12, 2013 9:06 am
sukhman wrote:can you plz make it clear why is not divided by 5 since 30=2*3*5 Though your answer is correct
Ideally, it should be done by considering for 2,3 and 5 and then consider the lowest power. However, we can see that "5" & "2" are present in almost all of the numbers.. So, the possibility of occurrence of "3" seems minimal; that's why I considered "3"
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