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# What is the value of N? (1) 10^N is the least positive integer which

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Joined: 22 Aug 2013
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08 Apr 2018, 05:11
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35% (medium)

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76% (01:41) correct 24% (02:35) wrong based on 86 sessions

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What is the value of N?

(1) 10^N is the least positive integer which is divisible by both 5^8 and 4^4.

(2) N is the least positive integer such that product of all integers from 1 to N, inclusive, is divisible by 128.
Manager
Joined: 30 Mar 2017
Posts: 136
GMAT 1: 200 Q1 V1
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08 Apr 2018, 07:39
I think it's D.

Statement 1
5^8 and 4^4 have to both divide 10^N, where N is minimized. We can find N (i.e. there exists a smallest N such that 10^N is divisible by both 5^8 and 4^4). Sufficient.

Statement 2
128 has to divide N!, where N is minimized. We can find N (i.e. there exists a smallest number N such that N! is div by 128). Sufficient.
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Joined: 19 Apr 2017
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What is the value of N? (1) 10^N is the least positive integer which  [#permalink]

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27 Sep 2018, 12:55
1
What is the value of N?

(1) $$10^N$$ is the least positive integer which is divisible by both $$5^8$$ and $$4^4.$$

$$10^N/(5^8*4^4)$$ can be written as $$10^N/(5^8*2^8) = 10^N/(10^8)$$ N has to be 8?

(2) N is the least positive integer such that product of all integers from 1 to N, inclusive, is divisible by 128.

$$128=2^7$$

N! has to contain at least 7 2's to be divisible by $$128 => N!/128 =N!/2^7$$

IF N = 7 then $$N! = {1*2*3*4*5*6*7} = {1*2*3*(2^2)*5*(2*3)*7}$$ this contains only 4 2's we need three more 2
if N = 8 then $$N! = {1*2*3*4*5*6*7*8} = {1*2*3*(2^2)*5*(2*3)*7*(2^3)}$$ this contains 7 twos so N= 8

Sufficient So D
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# What is the value of N? (1) 10^N is the least positive integer which

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