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36 distinct factors in S we need 2 numbers for a product, so 2C36 = 630 ?
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help needed.. Total 36 factors and selecting 2 from them gives a no. not in option.

Noshad
Let S be the set of all positive integer divisors of 100,000. How many numbers are the product of two distinct elements of S?

A. 98

B.100

C. 117

D. 119

E. 121
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Noshad
Let S be the set of all positive integer divisors of 100,000. How many numbers are the product of two distinct elements of S?

A. 98

B.100

C. 117

D. 119

E. 121

factors of 10^5 = 2^5*5^5; total factors = 36
i am not able to solve further this question GMATinsight ; sir please advise on the solution

How do you determine the total factors to be 36? Kindly explain..
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Archit3110
Noshad
Let S be the set of all positive integer divisors of 100,000. How many numbers are the product of two distinct elements of S?

A. 98

B.100

C. 117

D. 119

E. 121

factors of 10^5 = 2^5*5^5; total factors = 36
i am not able to solve further this question GMATinsight ; sir please advise on the solution

How do you determine the total factors to be 36? Kindly explain..

By prime factorising 1,00,000

1,00,000 can be written as 10^5
Further break 10^5 into (2*5)^5
2^5 * 5^5.. when you have a number in the form of a^p * b^q * c^r ..... The total number of factors of that number will be
(p+1)(q+1).....

So (5+1)(5+1) = 36..

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thyagi

By prime factorising 1,00,000

1,00,000 can be written as 10^5
Further break 10^5 into (2*5)^5
2^5 * 5^5.. when you have a number in the form of a^p * b^q * c^r ..... The total number of factors of that number will be
(p+1)(q+1).....

So (5+1)(5+1) = 36..

Posted from my mobile device

Oh yes! Thank you! Feel silly now.. :hurt:
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chetan2u, Gladiator59, Bunuel, VeritasKarishma, generis

dear experts :

Could you please help? Like others,I am also not able to solve this.

Regards,
Arup Sarkar
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Noshad
Let S be the set of all positive integer divisors of 100,000. How many numbers are the product of two distinct elements of S?

A. 98

B.100

C. 117

D. 119

E. 121


100,000 = 10^5 = 2^5 * 5^5

No of distinct factors = 6*6 = 36
These are: 1, 2^1, 2^2, 5, 2^3, 2^1*5^1, 2^4, 5^2, ...

When you multiply any two of these, you will get a product with powers of 2 and/or 5.
Since 2^5 and 5^5 are the maximum powers of 2 and 5, the power of either that you can get ranges from 0 to 10.

This gives us 11*11 = 121 numbers

But no number can be multiplied by itself so the largest and smallest powers of 2 and 5 will not multiply with themselves.

Hence, you will not get the following 4 numbers:

2^10 * 5^10 (because there is only one 2^5 * 5^5)

2^10 (because there is only one 2^5)

5^10 (because there is only one 5^5)

2^0*5^0 (because there is only one 2^0*5^0)

So in all, you will have 117 numbers.

Answer (C)
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VeritasKarishma
Noshad
Let S be the set of all positive integer divisors of 100,000. How many numbers are the product of two distinct elements of S?

A. 98

B.100

C. 117

D. 119

E. 121


100,000 = 10^5 = 2^5 * 5^5

No of distinct factors = 6*6 = 36
These are: 1, 2^1, 2^2, 5, 2^3, 2^1*5^1, 2^4, 5^2, ...

When you multiply any two of these, you will get a product with powers of 2 and/or 5.
Since 2^5 and 5^5 are the maximum powers of 2 and 5, the power of either that you can get ranges from 0 to 10.

This gives us 11*11 = 121 numbers

But no number can be multiplied by itself so the largest and smallest powers of 2 and 5 will not multiply with themselves.

Hence, you will not get the following 4 numbers:

2^10 * 5^10 (because there is only one 2^5 * 5^5)

2^10 (because there is only one 2^5)

5^10 (because there is only one 5^5)

2^0*5^0 (because there is only one 2^0*5^0)

So in all, you will have 117 numbers.

Answer (C)


HI VeritasKarishma

Applogies, but I am still not able to understand the solution that you posted. Please help.

chetan2u GMATinsight gmatbusters
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VeritasKarishma ; hello ma'am..
I am not able to understand from the highlighted part onwards..
a. how did 121 numbers come up?
b. 2^10 , 5^10 are the highest power but how did you deduce to 117 numbers :?

GMATinsight ; sir if possible kindly look into the question and assist.

VeritasKarishma
Noshad
Let S be the set of all positive integer divisors of 100,000. How many numbers are the product of two distinct elements of S?

A. 98

B.100

C. 117

D. 119

E. 121


100,000 = 10^5 = 2^5 * 5^5

No of distinct factors = 6*6 = 36
These are: 1, 2^1, 2^2, 5, 2^3, 2^1*5^1, 2^4, 5^2, ...

When you multiply any two of these, you will get a product with powers of 2 and/or 5.
Since 2^5 and 5^5 are the maximum powers of 2 and 5, the power of either that you can get ranges from 0 to 10.

This gives us 11*11 = 121 numbers

But no number can be multiplied by itself so the largest and smallest powers of 2 and 5 will not multiply with themselves.

Hence, you will not get the following 4 numbers:

2^10 * 5^10 (because there is only one 2^5 * 5^5)
2^10 (because there is only one 2^5)

5^10 (because there is only one 5^5)

2^0*5^0 (because there is only one 2^0*5^0)

So in all, you will have 117 numbers.

Answer (C)
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GMATinsight ; thanks for the solution :) , this surely is a unique question not sure whether is GMAT type or not as it involves a lot of combining of pairs of factors solving under 120 sec would be really challenging.. :( :shocked

Noshad ; whats the source of this question? is it really from gmat source ? I actually googled this question and it comes up to be a question from American mathematic association /olympiad, is it so?
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Why wouldn't 36C2 give the solution...since the question is "How many numbers are the product of two distinct elements of S?"

Won't the above solution be more suitable for "How many distinct numbers are the product of two distinct elements of S?"

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VeritasKarishma
Noshad
Let S be the set of all positive integer divisors of 100,000. How many numbers are the product of two distinct elements of S?

A. 98

B.100

C. 117

D. 119

E. 121


100,000 = 10^5 = 2^5 * 5^5

No of distinct factors = 6*6 = 36
These are: 1, 2^1, 2^2, 5, 2^3, 2^1*5^1, 2^4, 5^2, ...

When you multiply any two of these, you will get a product with powers of 2 and/or 5.
Since 2^5 and 5^5 are the maximum powers of 2 and 5, the power of either that you can get ranges from 0 to 10.

This gives us 11*11 = 121 numbers

But no number can be multiplied by itself so the largest and smallest powers of 2 and 5 will not multiply with themselves.

Hence, you will not get the following 4 numbers:

2^10 * 5^10 (because there is only one 2^5 * 5^5)

2^10 (because there is only one 2^5)

5^10 (because there is only one 5^5)

2^0*5^0 (because there is only one 2^0*5^0)

So in all, you will have 117 numbers.

Answer (C)

Hi, I am not sure whether I am getting the question right or I am missing anything.

I understood the solution posted, but I cannot understand why "If we have 36 factors and we are to select and 2 from these factors, then why is 36C2 wrong?

Can you please explain this?

Thanks

VeritasKarishma GMATinsight
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VeritasKarishma
Noshad
Let S be the set of all positive integer divisors of 100,000. How many numbers are the product of two distinct elements of S?

A. 98

B.100

C. 117

D. 119

E. 121


100,000 = 10^5 = 2^5 * 5^5

No of distinct factors = 6*6 = 36
These are: 1, 2^1, 2^2, 5, 2^3, 2^1*5^1, 2^4, 5^2, ...

When you multiply any two of these, you will get a product with powers of 2 and/or 5.
Since 2^5 and 5^5 are the maximum powers of 2 and 5, the power of either that you can get ranges from 0 to 10.

This gives us 11*11 = 121 numbers

But no number can be multiplied by itself so the largest and smallest powers of 2 and 5 will not multiply with themselves.

Hence, you will not get the following 4 numbers:

2^10 * 5^10 (because there is only one 2^5 * 5^5)

2^10 (because there is only one 2^5)

5^10 (because there is only one 5^5)

2^0*5^0 (because there is only one 2^0*5^0)

So in all, you will have 117 numbers.

Answer (C)

Hi, I am not sure whether I am getting the question right or I am missing anything.

I understood the solution posted, but I cannot understand why "If we have 36 factors and we are to select and 2 from these factors, then why is 36C2 wrong?

Can you please explain this?

Thanks

VeritasKarishma GMATinsight

The number has 36 factors such as 1, 2, 5, 10 ...

If you were to select any two factors of the 36 factors, 36C2 would be correct. But we need the number of distinct products of any two factors.
36C2 will involve a lot of double counting.
Say I select 2 and 5. Product = 10
Say I select 1 and 10. Product = 10
(Then they need to be taken as a single case only since the product is 10 in both cases.)

Similarly there will be many ways in which different products can be obtained using 2 of the 36 factors.
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but doesn't the problem say the following:

"how many numbers are the product of 2 DISTINCT ELEMENTS?"

therefore, if we got the same Result with 2 * 5 and 1 * 10 it does not matter according the question. We have used 2 DISTINCT ELEMENTS from the set.

Maybe the wording should be fixed?
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