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

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.

Thanks!

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

The solution presented below justifies the "do-this-in-this-situation" approach presented in some previous posts.

6,8, 10

FIND LCM:

/2 to make calculation easier:

3,4,5
3x4x5= 60

See answer that is the first multiple of 60 and add 1
60
120+1 = C

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