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MathRevolution
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MathRevolution
[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

Cheers,
Brent
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=>

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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