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E
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Difficulty:
95%
(hard)
Question Stats:
34% (02:19) correct
66%
(02:07)
wrong
based on 180
sessions
History
Date
Time
Result
Not Attempted Yet
What is the least possible number of 0s(zeroes) at the end of integer k, if \((0.0025)(0.025)(0.00025)×10*k\) is an integer and k is also an integer?
A) 12
B) 9
C) 8
D) 6
E) 5
A) 12
B) 9
C) 8
D) 6
E) 5
ENGRTOMBA2018
Joined: 20 Mar 2014
Last visit: 01 Dec 2021
Posts: 2,319
Given Kudos: 816
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE:Engineering (Aerospace and Defense)
Products:
15 Jan 2016, 08:06
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chetan2u
As we need to have both k and the expression (0.0025)(0.025)(0.00025)×10*k as integers, I can mention
(0.0025)(0.025)(0.00025)×10*k = I , where I is an integer. For minimum value of k, I =1.
\((1/4)^3*10^{-6}*10k = 1\)---> \(k = 64*10^5\). As k must be an integer, you can not express it as \(6.4*10^6\) or \(0.64*10^7\). Additionally, you can not write it as \(640*10^4\) etc as you need the MINIMUM value of k.
Thus 5 is the correct answer. E is the correct answer.
General Discussion
17 Jan 2016, 01:04
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chetan2u
Hi,
the equation given to us is ..
\((0.0025)(0.025)(0.00025)×10*k\) ...
if we get this in power of 10, it becomes \(0.25*10^{-2}*0.25*10^{-1}*0.25*10^{-3}*10*k\)..
= \(0.25^3*10^{-5}*k\)...
so for \(0.25^3*10^{-5}*k\) to be an integer \(k=4^3*10^5\)...
since we are looking for the number of 0s in k, k has 5 zeroes..
E
