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23 Aug 2018, 02:13
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
83% (00:58) correct
17%
(01:03)
wrong
based on 87
sessions
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Not Attempted Yet
[Math Revolution GMAT math practice question]
A whole number greater than \(1\) has remainders of \(1\) when it is divided by each of the numbers of \(2, 3, 4\) and \(5\). What is the smallest such number?
\(A. 31\)
\(B. 51\)
\(C. 61\)
\(D. 91\)
\(E. 121\)
A whole number greater than \(1\) has remainders of \(1\) when it is divided by each of the numbers of \(2, 3, 4\) and \(5\). What is the smallest such number?
\(A. 31\)
\(B. 51\)
\(C. 61\)
\(D. 91\)
\(E. 121\)
23 Aug 2018, 02:29
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MathRevolution
LCM(2,3,4,5)=60
The number can be written in the form 60k+1
so 60k+1 would be the smallest when k=1
So 60k+1=60*1+1=60+1=61
Ans .(C)
23 Aug 2018, 04:23
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MathRevolution
Strategy - Test The Answers
A - 31 - Rem>1 when divided by 4
B - 51 - Div. by 3
C - 61 - Leaves remainder 1 when div by 2,3,4,5
Answer C.
