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What is the smallest integer greater than 1 that leaves a remainder of

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New post 03 Dec 2015, 23:49
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What is the smallest integer greater than 1 that leaves a remainder of  [#permalink]

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New post 04 Dec 2015, 01:06
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Bunuel 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


hi,
whenever we have these type of Q, the first step is to find the LCM of the given number and add the remainder to it...
LCM of 6,8,10=120..
so integer=120+1=121
C
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Re: What is the smallest integer greater than 1 that leaves a remainder of  [#permalink]

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New post 06 Dec 2015, 06:30
Or u can just use the answer choices here. Since the answers are already arranged in ascending order, the first number which gives remainder as 1 for all three is the correct answer. In the given question, the first number which gives a remainder of 1 for 6,8 and 10 is 121.

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Re: What is the smallest integer greater than 1 that leaves a remainder of  [#permalink]

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New post 06 Dec 2015, 09:13
Bunuel 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


This problem can be solve in a 2 step process

1. Find the Smalles number divisible by 6, 8, and 10 ( LCM of 6, 8, and 10 )
2. Add 1 to the result of step no 1

1. LCM of 6, 8, and 10 is 120
2. Smallest number + 1 = 121

So the answer is definitely (C) 121
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Re: What is the smallest integer greater than 1 that leaves a remainder of  [#permalink]

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New post 10 Aug 2017, 09:47
Bunuel 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


We need to determine the smallest integer greater than 1 that leaves a remainder of 1 when divided by any of the integers 6, 8, and 10.

Let’s analyze each answer choice.

Since 21/6 has a reminder of 3, answer A is not correct.

Since 41/6 has a remainder of 5, answer B is not correct.

Since 121/6, 121/8, and 121/10 all produce a remainder of 1, answer C is correct.

Alternate Solution:

If x is the smallest number greater than 1 that leaves a remainder of 1 when divided by 6, 8, and 10, then x - 1 must be the smallest number that is divisible by 6, 8, and 10. So, let’s find the LCM of 6, 8, and 10. Since 6 = 3 x 2, 8 = 2^3, and 10 = 2 x 5, LCM(6, 8, 10) = 2^3 x 3 x 5 = 120.

Since x - 1 = 120, x must equal 121.

Answer: C
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What is the smallest integer greater than 1 that leaves a remainder of  [#permalink]

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New post 23 Aug 2018, 07:04
The solution presented below justifies the "do-this-in-this-situation" approach presented in some previous posts.
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What is the smallest integer greater than 1 that leaves a remainder of &nbs [#permalink] 23 Aug 2018, 07:04
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