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Problem Solving (PS) | Re: A and B are two multiples of 36, and Q is the set of consecutive integ

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MathRevolution wrote:

Lengthy Method:

Multiple of 36: 36, 72, 108, 144...

Q = 72 to 144 [9 multiples of 9: 72, 81, 90, 99, 108, 117, 126, 135, 144]

From 72 to 144 inclusive, multiples of 4 will be:

144 = 72 + (n-1) * 4

=> 72 = 4n - 4

=> n = 19

Answer B


Shortcut:

Between every two consecutive multiples of 36, there will be ten multiple of 4

72 = 36 + (n-1) * 4

=> 36 = 4n - 4

=> n = 10


Therefore, between two multiples of 36 (36 and 72) lies ten multiple of 4

So if there are 9 multiples of 9 in Q, the number of multiples of 4 would be = 9 + 10 = 19

Answer B

­WHY 0 is not considered a multiple of 36 here? 0 is multiple of all numbers.

Statistics : Posted by mithunkrishna • on 18 May 2021, 05:30 • Replies 4 • Views 2679



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