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There are 3 ways to make the number 12 using products of two positive 24 Jul 2018, 01:26
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There are 3 ways to make the number 12 using products of two positive integers. These are 1*12, 2*6, and 3*4. In how many ways can 2700 be written as the product of two positive integers?

A. 6
B. 12
C. 15
D. 18
E. 36
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D
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There are 3 ways to make the number 12 using products of two positive 24 Jul 2018, 04:28
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MathRevolution
[Math Revolution GMAT math practice question]

There are 3 ways to make the number 12 using products of two positive integers. These are 1*12, 2*6, and 3*4. In how many ways can 2700 be written as the product of two positive integers?

A. 6
B. 12
C. 15
D. 18
E. 36
Show more


Most of us by now know how to find factors. But even after knowing this, you can go wrong if you miss out on "pair of factors" and mark "number of factors"

1) Prime factorise 2700..
\(2700=3^3*10^2=3^3*2^2*5^2\)
2) number of factors ..
(3+1)(2+1)(2+1)=4*3*3=36..

But we are looking for pairs so 35/2=18 as product of two factors say 1*2700=10*270=2*1350...

Some extra points
If a number is a square, the number of factors will be ODD
For example say it was 8100..
\(8100=3^4*2^2*5^2.....5*3*3=45\)
So now 45/2 is not an integer...so what does one do..
Two cases..
1) if question asks you pair of factors then add 1 as \(90*90 =8100\)
So \(\frac{(45+1)}{2}=23\)
2) but if question asks you pair of different factors then subtract one as 90*90 will not be included so \((\frac{45-1)}{2}=22\)
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There are 3 ways to make the number 12 using products of two positive 24 Jul 2018, 02:24
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MathRevolution
[Math Revolution GMAT math practice question]

There are 3 ways to make the number 12 using products of two positive integers. These are 1*12, 2*6, and 3*4. In how many ways can 2700 be written as the product of two positive integers?

A. 6
B. 12
C. 15
D. 18
E. 36
Show more

2700 = (2^2 × 3^3 × 5^2)
total number of factors are (2+1)(3+1)(2+1)=36 including 1 and itself
no of ways 2700 can be written as product of two positive int = 36/2=18
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There are 3 ways to make the number 12 using products of two positive Updated on: 02 May 2021, 05:42
Originally posted by GMATGuruNY on 24 Jul 2018, 03:57.
Last edited by GMATGuruNY on 02 May 2021, 05:42, edited 1 time in total.
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MathRevolution
[Math Revolution GMAT math practice question]

There are 3 ways to make the number 12 using products of two positive integers. These are 1*12, 2*6, and 3*4. In how many ways can 2700 be written as the product of two positive integers?

A. 6
B. 12
C. 15
D. 18
E. 36
Show more

To count the factors of a positive integer:
1. Prime-factorize the integer
2. Write the prime-factorization in the form \((a^p)(b^q)(c^r)\)...
3. The number of factors = \((p+1)(q+1)(r+1)\)...

\(2700 = 2^2 * 3^3 * 5^2\)

Adding 1 to each exponent and multiplying, we get:
Total number of factors = \((2+1)(3+1)(2+1) = 36\)

These 36 factors can be used to form 18 FACTOR PAIRS, as follows:
1*2700
2*1350
3*900
And so on.

Show Spoiler
D
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There are 3 ways to make the number 12 using products of two positive 24 Jul 2018, 10:08
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chetan2u
MathRevolution
[Math Revolution GMAT math practice question]

There are 3 ways to make the number 12 using products of two positive integers. These are 1*12, 2*6, and 3*4. In how many ways can 2700 be written as the product of two positive integers?

A. 6
B. 12
C. 15
D. 18
E. 36


Most of us by now know how to find factors. But even after knowing this, you can go wrong if you miss out on "pair of factors" and mark "number of factors"

1) Prime factorise 2700..
\(2700=3^3*10^2=3^3*2^2*5^2\)
2) number of factors ..
(3+1)(2+1)(2+1)=4*3*3=36..

But we are looking for pairs so 35/2=18 as product of two factors say 1*2700=10*270=2*1350...

Some extra points
If a number is a square, the number of factors will be ODD
For example say it was 8100..
8100=3^4*2^2*5^2.....5*3*3=45
So now 45/2 is not an integer...so what does one do..
Two cases..
1) if question asks you pair of factors then add 1 as 900*900 =8100
So (45+1)/2=23
2) but if question asks you pair of different factors then subtract one as 900*900 will not be included so (45-1)/2=22
Show more

Hi chetan2u,
I the question asks about the sum of the factors then how to go about that?
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There are 3 ways to make the number 12 using products of two positive 26 Jul 2018, 01:45
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=>

\(2700 = 2^2*3^3*5^2\)
The number of distinct factors of \(2700\) is \((2+1)(3+1)(2+1) = 36.\)
Since the order of multiplication does not matter (i.e. \(30 * 90 = 90*30\)), the number of pairs of positive integers that multiply to give \(2700\) is \(\frac{36}{2} = 18.\)

Therefore, the answer is D.
Answer: D
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There are 3 ways to make the number 12 using products of two positive 27 Jul 2018, 23:37
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chetan2u
MathRevolution
[Math Revolution GMAT math practice question]

There are 3 ways to make the number 12 using products of two positive integers. These are 1*12, 2*6, and 3*4. In how many ways can 2700 be written as the product of two positive integers?

A. 6
B. 12
C. 15
D. 18
E. 36


Most of us by now know how to find factors. But even after knowing this, you can go wrong if you miss out on "pair of factors" and mark "number of factors"

1) Prime factorise 2700..
\(2700=3^3*10^2=3^3*2^2*5^2\)
2) number of factors ..
(3+1)(2+1)(2+1)=4*3*3=36..

But we are looking for pairs so 35/2=18 as product of two factors say 1*2700=10*270=2*1350...

Some extra points
If a number is a square, the number of factors will be ODD
For example say it was 8100..
8100=3^4*2^2*5^2.....5*3*3=45
So now 45/2 is not an integer...so what does one do..
Two cases..
1) if question asks you pair of factors then add 1 as 900*900 =8100
So (45+1)/2=23
2) but if question asks you pair of different factors then subtract one as 900*900 will not be included so (45-1)/2=22

Hi chetan2u,
I the question asks about the sum of the factors then how to go about that?
Show more

hi..
not likely you will such a question in actual, it will limit to number of factors and not sum of factors ..
but formula for it is ..
1) Prime factorise 2700..
\(2700=3^3*10^2=3^3*2^2*5^2\)
    a) number of factors ..
    \((3+1)(2+1)(2+1)=4*3*3=36..\)
    b) sum of factors ..
    for \(a^x*b^y*c^z\), it is \(\frac{a^{(x+1)}-1}{a-1}*\frac{b^{(y+1)}-1}{b-1}*\frac{c^{(z+1)}-1}{c-1}\)..
    \(\frac{2^{(2+1)}-1}{2-1}*\frac{3^{(3+1)}-1}{3-1}*\frac{5^{(2+1)}-1}{5-1}=7*\frac{3^4-1}{2}*\frac{5^3-1}{4}=7*40*31=8680\)
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There are 3 ways to make the number 12 using products of two positive 02 May 2021, 04:50
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better explanation


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There are 3 ways to make the number 12 using products of two positive 02 May 2021, 23:09
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For a Non-Perfect Square Number, the number of ways to have 2 positive integer multiply to equal that number is = the number of “factor pairs” you would find if you made a Factor Tree with all the pairs of factors.

This is simply = (Total Number of Unique positive Factors) * (1/2)

2,700 = (2)^2 * (3)^3 * (5)^2

Total number of unique positive factors = 3 * 4 * 3 = 36

36 * (1/2) = 18

There are 18 unordered ways to have 2 positive integers multiply to equal 2,700

(1 and 2,700)

(2 and 1,350)

Etc.

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There are 3 ways to make the number 12 using products of two positive 22 Nov 2025, 05:18
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