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Let S be the set of all positive integer divisors of 100,000. How many

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Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

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18 Mar 2019, 08:02
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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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Re: Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

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26 Mar 2019, 09:39
1
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
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Re: Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

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29 Mar 2019, 04:00
36 distinct factors in S we need 2 numbers for a product, so 2C36 = 630 ?
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Re: Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

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31 Mar 2019, 21:16
help needed.. Total 36 factors and selecting 2 from them gives a no. not in option.

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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Location: India
GMAT 1: 700 Q49 V36
GPA: 4
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Re: Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

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31 Mar 2019, 23:29
Archit3110 wrote:
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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Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

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01 Apr 2019, 00:20
3
OhsostudiousMJ wrote:
Archit3110 wrote:
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..

Posted from my mobile device
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Re: Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

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01 Apr 2019, 00:28
thyagi wrote:

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..
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Re: Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

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04 Apr 2019, 11:14

dear experts :

Regards,
Arup Sarkar
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Re: Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

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05 Apr 2019, 02:03
2
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.

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Re: Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

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06 Apr 2019, 23:53
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.

chetan2u GMATinsight gmatbusters
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Re: Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

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07 Apr 2019, 01:02
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.

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.

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Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

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Updated on: 07 Apr 2019, 02:51
2
1
Archit3110 wrote:
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.

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.

Calculation of Number of Factors
For a Number $$N=a^p*b^q*c^r*...$$where a, b, c etc are distinct prime factors of n and p, q r are their respective exponents
The number of factors of $$N = (p+1)*(q+1)*(r+1)*...$$ and so on

i.e. $$100,000 = 10^5 = 2^5 * 5^5$$

hence, total factors of 100000 = (5+1)*(5+1) = 36

i.e. S = {1, $$2^1$$, $$2^2$$, $$5^1$$, $$2^3$$, $$2^1*5^1$$, $$2^4$$, $$2^2*5$$........100000} 36 elements in S

The biggest terms of the set are $$2^5*5^5$$ and $$2^4*5^5$$ and $$2^5*5^4$$

i.e. the biggest product of two terms in set S will be $$2^5*5^5$$*$$2^4*5^5$$ and $$2^5*5^5$$*$$2^5*5^4$$

i.e. Biggest terms of the product will be $$2^9*5^{10}$$ and $$2^{10}*5^9$$

i.e. $$2^0$$, $$2^1$$, $$2^2$$--- $$2^{10}$$ etc. will have combination with $$5^0$$ i.e. 11 combinations
Similarly, i.e. $$2^0$$, $$2^1$$, $$2^2$$--- $$2^{10}$$ etc. will have combination with $$5^1$$ i.e. 11 combinations
Similarly, i.e. $$2^0$$, $$2^1$$, $$2^2$$--- $$2^{10}$$ etc. will have combination with $$5^2$$ i.e. 11 combinations
and so on...

i.e. $$2^0$$, $$2^1$$, $$2^2$$--- $$2^{10}$$ etc. will have combination with $$5^{10}$$ i.e. 11 combinations

i.e. Total combinations = 11*11 = 121

But as already mentioned the combinations that are not possible but included here are $$2^{10}*5^{10}$$ and $$2^0*5^0$$ and $$2^{10}$$ and $$5^{10}$$

Total Acceptable combinations = 121 - 4 = 117

Archit3110 warrior1991
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Originally posted by GMATinsight on 07 Apr 2019, 02:06.
Last edited by GMATinsight on 07 Apr 2019, 02:51, edited 2 times in total.
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Let S be the set of all positive integer divisors of 100,000. How many  [#permalink]

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07 Apr 2019, 02:38
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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..

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?
Let S be the set of all positive integer divisors of 100,000. How many   [#permalink] 07 Apr 2019, 02:38
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