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What is the smallest integer greater than 1

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by Vincen » Wed Sep 20, 2017 7:00 am
What is the smallest integer greater than 1 that leaves a remainder of 1 when divided by any of the integers 6, 8, and 10?

A) 21
B) 41
C) 121
D) 241
E) 481

The OA is C.

How can I get the correct answer? I don't know how to can I do it.
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What is the smallest integer greater than 1

by Brent@GMATPrepNow » Wed Sep 20, 2017 7:27 am
Vincen wrote:What is the smallest integer greater than 1 that leaves a remainder of 1 when divided by any of the integers 6, 8, and 10?

A) 21
B) 41
C) 121
D) 241
E) 481
The remainder is 1 when the integer is divided by 6
Check the answer choices...
A) 21 divided by 6 equals 3 with remainder 3. ELIMINATE A
B) 41 divided by 6 equals 6 with remainder 5. ELIMINATE B
C) 121 divided by 6 equals 20 with remainder 1. KEEP C
D) 241 divided by 6 equals 40 with remainder 1. KEEP D
C) 481 divided by 6 equals 80 with remainder 1. KEEP E

The remainder is 1 when the integer is divided by 8
C) 121 divided by 8 equals 15 with remainder 1. KEEP C
D) 241 divided by 8 equals 30 with remainder 1. KEEP D
C) 481 divided by 8 equals 60 with remainder 1. KEEP E

The remainder is 1 when the integer is divided by 10
C) 121 divided by 10 equals 1 with remainder 1. KEEP C

Since the question asks for the SMALLEST integer, and since we've already eliminated A and B (numbers that are both smaller than 121), we can be certain that the correct answer is C

RELATED VIDEO: https://www.youtube.com/watch?v=XCxgIR042tE

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Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Matt@VeritasPrep » Wed Sep 20, 2017 4:48 pm
You could also find the LCM of 6, 8, and 10, then add 1 to it. :)

This trick generalizes well to other such remainder problems too!
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by Scott@TargetTestPrep » Mon Sep 25, 2017 3:33 pm
Vincen wrote:What is the smallest integer greater than 1 that leaves a remainder of 1 when divided by any of the integers 6, 8, and 10?

A) 21
B) 41
C) 121
D) 241
E) 481
The key to solving this problem is to find the least common multiple (LCM) of 6, 8, and 10, since the LCM of these three numbers will leave a remainder of 0 when divided by each of them. Since the LCM of 6, 8, and 10 is 120, the smallest number that leaves a remainder of 1 when divided by any of these numbers is 121.

Alternate Solution:

We can try each answer choice starting from the smallest one. We see that answer choice A does not meet the given criteria, since 21 divided by 6 leaves a remainder of 3. We see that answer choice B does not meet the given criteria either, since 41 divided by 6 leaves a remainder of 5. Since 121 leaves a remainder of 1 when divided by 6, 8, and 10, and since we are looking for the smallest such number, that is the number we are looking for.

Answer: C

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