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

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

by Max@Math Revolution » Thu Aug 23, 2018 1:08 am

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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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by GMATGuruNY » Thu Aug 23, 2018 3:05 am
Max@Math Revolution 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
We can PLUG IN THE ANSWERS.
To yield a remainder of 1 when divided by 4, the correct answer must be 1 more than a multiple of 4.
The smallest answer choice that is 1 more than a multiple of 4 is 61.
When 61 is divided by 2, 3 or 5, the remainder is 1 in each case.

The correct answer is C.
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by fskilnik@GMATH » Thu Aug 23, 2018 4:34 am
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by fskilnik@GMATH » Thu Aug 23, 2018 7:11 am
In the solution I posted above, I should have started with N>1 int (in the first line).
Please note that N=1 would be an acceptable answer IF the question stem would allow it... (It does not, of course.)
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by Brent@GMATPrepNow » Thu Aug 23, 2018 7:18 am
Max@Math Revolution 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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by Max@Math Revolution » Sun Aug 26, 2018 5:19 pm
=>

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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