If the prime numbers p and t are the only prime factors of

If the prime numbers p and t are the only prime factors of the integer m, is m a multiple of p²t?


We are told that \(p\) and \(t\) are the ONLY prime factors of m. It could be expressed as \(m=p^x*t^y\), where \(x\) and \(y\) are integers \(\geq{1}\).

Question: is \(m\) a multiple of \(p^2*t\). We already know that \(p\) and \(t\) are the factors of \(m\), so basically question asks whether the power of \(p\), in our prime factorization (denoted as \(x\)) is more than or equal to 2: so is \(x\geq{2}\).

(1) m has more than 9 positive factors:

Formula for counting the number of distinct factors of integer \(n\) expressed by prime factorization as: \(n=a^x*b^y*c^z\), is \((x+1)(y+1)(z+1)\). This also includes the factors 1 and \(n\) itself.

We are told that \((x+1)(y+1)>9\) (since \(m=p^x*t^y\))
But it's not sufficient to determine whether \(x\geq{2}\). (\(x\) can be 1 and \(y\geq{4}\) and we would have their product \(>9\), e.g. \((1+1)(4+1)=10\).) Not sufficient.


(2) m is a multiple of p^3
This statement clearly gives us the value of power of \(p\), which is 3, \(x=3>2\). So \(m\) is a multiple of \(p^2t\). Sufficient.

Answer: B.
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Excellent explanation but are we assuming that 'p' and 't' are different prime factors i.e. 'p' is not equal to 't'?

Yes. I think from the stem we can get this.
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Excellent question as well as explanation.

thanks bunuel good expalanation

Ans is B , good one Bunuel
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Hey
Lets look at statement 1
m has more than 9 factors
Now if p and t are the only prime factors then the other factors would be a combination of p and t either with each other or with themselves.
Now among those 9 factors, the following 2 things could happen.
1. 2 factors would be 1 and m. The other factors could be \(p, t, t^2, t^3, t^4, t^5, t^6\). In this case the integer m is NOT a multiple of\(p^2t\).
2. The other seven factors could have\(p^2\). In that case m would be a multiple of \(p^2t\)
So, Insufficient.
Lets look at statement 2
If m is a multiple of\(p^3\), then m must be a multiple of \(p^2\). We know that m is already a multiple of t. So m must be a multiple of \(p^2t\).
Hence Sufficient.

Hope this helps.
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We can write \(m=p^a\cdot{t^b}\) for some positive integers \(a\) and \(b.\)

(1) The number of positive factors of \(m\) is \((a+1)(b+1)>9.\)
If \(a=1\) and \(b>3\) then \(m=pt^b\) is not a multiple of \(p^2t.\)
If \(a>1\) then the answer is yes.
Not sufficient.

(2) Obviously sufficient.

Answer B.
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I got your second statement, but somehow I am not able to get the 1st statement.

For example you have p=2 and t=3, two prime numbers . Now the other 7 numbers can be any positive integer right ? i.e 4,6,8,9,4,6,8 isnt it ?

And the second case maybe that we have other 7 factors that include 2 and 3 as well . ex. 2,3,2,3,2,3,3,2,2 . In this case m is a multiple of p^2*t .

Is my thinking right ?

The formula you've written - (a+1)(b+1) is for the no of prime factors of a number right ? .

NO. It is for all the positive factors of the number, including 1 and the number itself, not only prime factors.
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Check this: math-number-theory-88376.html It might help.
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Note that these factors are combinations of powers of the prime factors only.
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Thanks Bunnuel ! I have gone through that, very valuable !

m = p^x * t^y where x is at least 1 and y is at least 1...
For m to be a multiple of p^2 * t then m must have at least 2 p and at least 1 t...

1. m has more than 9 factors
If m = p^1 * t^4 => number of factors = (1+1)(4+1) = 10 NOT A MULTIPLE!
If m = p^2 * t^3 => numbr of factors = (2+1)(3+1) = 12 A MULTIPLE!
INSUFFICIENT!

2. m is a multiple of p^3
Is it at least 2 factors of p? According to Statement (2) - YES!
Is it at least 1 factor of t? According to GIVEN - YES!

SUFFICIENT!

ANswer: B

IS my translation for this problem correct the given info...

we know that \(\frac{m}{p*t}\) = Integer since p and t are different integers

The question is now framed as is\(\frac{M}{p^2 t}\) ?( T is irrelevant for this question )

Yes, the question asks whether m/(p^2t)=integer, while saying that m/(pt)=integer.
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I could answer the question in 1.5 min. But I have never known the formula for counting the number of distinct factors of a integer. Thanks a ton, Bunuel

If the prime numbers p and t are the only prime factors of the integer m, is m a multiple of t*p^2?
(1) m has more than 9 positive factors
(2) m is a multiple of p^3

Draw a prime box and put p and t inside. According to the problem, there could be multiple instances of p and t in there, but that's it. We want to know whether there are at least two p's and one t in there.

Start with statement 2. If m is a multiple of p^3, that means there are 3 p's in m's prime box. There's already a t in there, according to the original question. So there are at least 2 p's and one t. Answer to question is yes, so statement is sufficient. Eliminate A, C, E.

Statement 1. Notice that this just says "positive factors" NOT prime factors. The complete set of factors is made by multiplying the prime factors in different combinations. For example, 12 has the prime factors 2, 2, and 3. We can find all of the general factors of 12 by taking 2, 3, 2*2, 2*3, 2*2*3, and of course 1.

So m has more than 9 positive factors. Well, I know m has p and t - there are 2 factors. And I know m has 1 and itself - there are 2 more factors, for a total of four. I need five more, so I have to add to my prime box to be able to create five more general factors. The only things I can put in my prime box are p and t. I can put all p's, all t's, or some combination of p's and t's. If I put in at least one p, then I'd have at least 2 p's and one t, which would answer the question "yes." BUT, if I put in all t's, then I'd only have one p, which would answer the question "no" - so the statement is insufficient.

Hence B.
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