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Solution:

In such a question, first find the number that is perfectly divisible by 3,5,7 and the smallest such number is the LCM of 3,5,7 = 105

However, we need a number which leaves a remainder of 2 !

=> We are not looking for the perfectly divisible 105 but 105 + 2 = 107 (option d)

Alternately, you can use the options and see if they leave remainder 2 when divided with 3,5 and 7.

Keep studying ! :student_woman:

Devmitra Sen
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38/3 also leaves a reminder of 2. And 38 is the smallest number which gives a R of 2. So shouldn't B be the answer? Please correct my thinking.
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Solution:

In such a question, first find the number that is perfectly divisible by 3,5,7 and the smallest such number is the LCM of 3,5,7 = 105

However, we need a number which leaves a remainder of 2 !

=> We are not looking for the perfectly divisible 105 but 105 + 2 = 107 (option d)

Alternately, you can use the options and see if they leave remainder 2 when divided with 3,5 and 7.

Keep studying ! :student_woman:

Devmitra Sen
GMAT SME

38/3 also leaves a reminder of 2. And 38 is the smallest number which gives a R of 2. So shouldn't B be the answer? Please correct my thinking.
Hi AbhaGanu,
We should get a remainder of 2 when divided by each of the numbers, 3, 5 and 7.
Dividing 38/7 or 38/5, doesn’t give you a remainder of 2. Hence, it is not the correct answer!

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The common remainder is 2 after dividing by 3, 5, and 7. hence, divide by LCM (3,5,7) which is 105.

We need remainder as 2, therefore, 105 + 2 = 107.

Answer D
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Can you please explain why option B is wrong.
It always leave a remainder of 2.
38/3=12, 12/5=2, 2/7 = remainder of 2

Thanks


MathRevolution
The common remainder is 2 after dividing by 3, 5, and 7. hence, divide by LCM (3,5,7) which is 105.

We need remainder as 2, therefore, 105 + 2 = 107.

Answer D
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