M12-36

Official Solution:

How many positive integers less than 200 are either multiples of 13 or multiples of 12, but not multiples of both?

A. 28
B. 29
C. 30
D. 31
E. 32


The number of multiples of an integer in a range is given by the formula \(\frac{\text{last multiple in the range - first multiple in the range}}{\text{multiple}}+1\). Hence:

• Number of multiples of 13 in the given range \(\frac{(\text{last-first})}{\text{multiple}}+1=\frac{195-13}{13}+1=15\);

• Number of multiples of 12 in the given range \(\frac{(\text{last-first})}{\text{multiple}}+1=\frac{192-12}{12}+1=16\);

• Number of multiples of both 13 and 12 is 1, which is \(13*12=156\).

Note that the number 156 is a multiple of both 13 and 12 and is included in the count of multiples of 13 and the count of multiples of 12. Since we don't want it to be counted at all, we need to subtract it once from the total count of multiples of 13, and once from the total count of multiples of 12.

Therefore, the number of positive integers that are either multiples of 13 or multiples of 12, but not multiples of both, in the given range is \((15-1)+(16-1)=29\).


Answer: B
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I think this is a high-quality question and I agree with explanation. Great Q.

Binit.

I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with explanation

Dear Bunuel,

Is there a efficient way to find the last number in any given range that divides by an integer x.

In this question, for example, how do I figure out 195 is the last multiple of 13 when range is (0,200) faster?

Regards
Srinath

An in integer to be a multiple of 3, the sum of its digit must be a multiple of 3. For example, 315 is divisible by 3 because 3 + 1 + 5 = 9 and 9 is divisible by 3. In your example, 201 is a multiple of 3 (2 + 0 + 1 = 3), thus so must be 201 - 3 = 198.
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I think this is a high-quality question and I agree with explanation.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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I think this is a high-quality question and I agree with explanation.
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why are we subtracting 13*12 2 times> shouldnt we subtract it only once?

15+16-1

156, which is a multiple of both 13 and 12, is counted twice in both the multiples of 13 and the multiples of 12. Since we don't want it to be counted at all, we should subtract it twice.
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I think this is a high-quality question and I agree with explanation. I like the solution+explanation. However, the site has saved 28 as the correct answer instead of the reasonable solution provided in the explanation (B. 29) (16-1)+(15-1).