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Re: QOTD#13 The integer m is a multiple of 154, 250, and 264. [#permalink]
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yes.

However, the time allotted is on average.

You could take a bit more if the question is daunting and difficult.

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QOTD#13 The integer m is a multiple of 154, 250, and 264. [#permalink]
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sandy wrote:
The integer m is a multiple of 154, 250, and 264. Which of the following do NOT have to be
factors of m?
Indicate all possible values

A. 176
B. 242
C. 275
D. 924
E. 2,500
F. 7,000



Drill 3
Question: 12
Page: 290



Show: :: OA
A,B,E


A very straight forward answer.

if m is a multiple of 154, 250, & 256, it means that m is divisible by 154, 250, 264. So;

\(m/154\), \(m/ 250\), \(m/264\) will yield a remainder of zero.

This means that all the prime factors of each of the denominators are in the numerator m.

Prime factors of 154 = 2 * 7 * 11
Prime factors of 250 = 2 * 5 * 5 * 5
Prime factors of 264 = 2 * 2 * 2 * 3 * 11

So, m at least have the following prime factors.

Three 2s, one 3, three 5s, one 7, and one 11.

Now check each option.

Option A: prime factors of 176 -> 2,2,2,2,11 ----> We do not have four 2s in m. Therefore 176 can't be a factor of m. Correct

Option B: prime factors of 242 -> 2,11,11 ----> Not a factor. Correct

Option C: prime factors of 275 -> 5,5,11 ---> Factor. Incorrect

Option D: prime factors of 924 -> 2,2,3,7,11 ---> Factor. Incorrect

Option E: prime factors of 2500 -> 2,2,5,5,5,5 ---> Not Factor. m doesn't have four 5s. Correct

Option F: prime factors of 7000 -> 2,2,2,5,5,5,7 ---> Factor. Incorrect
Show: ::

Answer: A, B, E
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Re: QOTD#13 The integer m is a multiple of 154, 250, and 264. [#permalink]
I think there's an error with the solution here, please correct me if I'm wrong.
If we consider m to be a product of 154, 250 and 264 i.e. 10164000.

It'd still be considered a multiple of all 3 numbers and in that case 176 and 242 are perfectly dividing 10164000, but 2500 isn't. So then the answer should be only Option E.

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Re: QOTD#13 The integer m is a multiple of 154, 250, and 264. [#permalink]
achal wrote:
I think there's an error with the solution here, please correct me if I'm wrong.
If we consider m to be a product of 154, 250 and 264 i.e. 10164000.

It'd still be considered a multiple of all 3 numbers and in that case 176 and 242 are perfectly dividing 10164000, but 2500 isn't. So then the answer should be only Option E.

Posted from my mobile device


First off, 2500 is a divisor of 10164000. So, with your approach, there are no correct answers.

Also, multiplying all three values together will yield incorrect conclusions.
To see why, let's say the three numbers are 2, 3 and 6.
If we multiply them all together we get m = 36, which means 18 must be a factor of m.
However, m could also equal 6 (a much smaller number than 36). In this case, 18 is NOT a factor of m.
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Re: QOTD#13 The integer m is a multiple of 154, 250, and 264. [#permalink]
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