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A whole number greater than 1 has remainders of 1 when it is divided b

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A whole number greater than 1 has remainders of 1 when it is divided b  [#permalink]

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New post 23 Aug 2018, 01:13
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[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?

\(A. 31\)
\(B. 51\)
\(C. 61\)
\(D. 91\)
\(E. 121\)

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A whole number greater than 1 has remainders of 1 when it is divided b  [#permalink]

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New post 23 Aug 2018, 01:29
MathRevolution wrote:
[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?

\(A. 31\)
\(B. 51\)
\(C. 61\)
\(D. 91\)
\(E. 121\)


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)
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Re: A whole number greater than 1 has remainders of 1 when it is divided b  [#permalink]

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New post 23 Aug 2018, 03:23
MathRevolution wrote:
[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?

\(A. 31\)
\(B. 51\)
\(C. 61\)
\(D. 91\)
\(E. 121\)


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.
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Re: A whole number greater than 1 has remainders of 1 when it is divided b  [#permalink]

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New post 23 Aug 2018, 07:19
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MathRevolution wrote:
[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?

\(A. 31\)
\(B. 51\)
\(C. 61\)
\(D. 91\)
\(E. 121\)



Let N = the number in question.

If N divided by 2 leaves a remainder of 1, then N is 1 greater than some multiple of 2.
This means that N - 1 must be a multiple of 2.

Likewise, if N divided by 3 leaves a remainder of 1, then N is 1 greater than some multiple of 3.
This means that N - 1 must be a multiple of 3.

Etc...

So, N - 1 must be a multiple of 2, 3, 4 and 5
Since we're looking for the smallest possible value of N, we must find the LEAST common multiple of 2, 3, 4 and 5
The LEAST common multiple of 2, 3, 4 and 5 is 60

So, N - 1 = 60, which means N = 61

Answer: C

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Brent
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Re: A whole number greater than 1 has remainders of 1 when it is divided b  [#permalink]

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New post 26 Aug 2018, 17:23
=>

Let \(x\) be the smallest number satisfying the original condition.
Then \(x – 1\) is the least common multiple of \(2, 3, 4\) and \(5\).
So, \(x – 1 = 60.\)
Thus, \(x = 61\).

Therefore, the answer is C.
Answer: C
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Re: A whole number greater than 1 has remainders of 1 when it is divided b &nbs [#permalink] 26 Aug 2018, 17:23
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