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factors

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Manager
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factors [#permalink]

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New post 09 Nov 2010, 12:33
when 777 is divided by positive integer n, the remainder is 77. How many possibilities are there for n?

a. 2
b. 3
c. 4
d. 5
e. 6

Let me your thought process on this. I know we have to put this in remainder format of n(y) + 77= 777 so n(y)=700 but from here I would find the number of factors for 700 to be 18. Is there a quicker way of figuring out the number of factors greater than 77?

Kudos [?]: 97 [0], given: 7

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Re: factors [#permalink]

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New post 09 Nov 2010, 12:47
gettinit wrote:
when 777 is divided by positive integer n, the remainder is 77. How many possibilities are there for n?

a. 2
b. 3
c. 4
d. 5
e. 6

Let me your thought process on this. I know we have to put this in remainder format of n(y) + 77= 777 so n(y)=700 but from here I would find the number of factors for 700 to be 18. Is there a quicker way of figuring out the number of factors greater than 77?


Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).

So we'd have: \(777=qn+77\), where \(remainder=77<n=divisor\) --> \(qn=700=2^2*5^2*7\) --> as \(n\) must be more than 77 then \(n\) could take only 5 values: 100, 140, 175, 350, and 700 (the least factor more than 77 is \(2^2*5^2=100\), now if you substitute 2, 2^2, and 5 by 7 you'll get the factors more than 100 plus if you include 7 you'll get one more factor, 700 itself, so total 1+3+1=5).

Answer: D.

Discussed here: remainder-101074.html?hilit=possibilities#p782422

Hope it's clear.
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Kudos [?]: 132578 [0], given: 12326

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Re: factors [#permalink]

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New post 09 Nov 2010, 14:47
Bunuel wrote:
gettinit wrote:
when 777 is divided by positive integer n, the remainder is 77. How many possibilities are there for n?

a. 2
b. 3
c. 4
d. 5
e. 6

Let me your thought process on this. I know we have to put this in remainder format of n(y) + 77= 777 so n(y)=700 but from here I would find the number of factors for 700 to be 18. Is there a quicker way of figuring out the number of factors greater than 77?


Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).

So we'd have: \(777=qn+77\), where \(remainder=77<n=divisor\) --> \(qn=700=2^2*5^2*7\) --> as \(n\) must be more than 77 then \(n\) could take only 5 values: 100, 140, 175, 350, and 700 (the least factor more than 77 is \(2^2*5^2=100\), now if you substitute 2, 2^2, and 5 by 7 you'll get the factors more than 100 plus if you include 7 you'll get one more factor, 700 itself, so total 1+3+1=5).

Answer: D.

Discussed here: remainder-101074.html?hilit=possibilities#p782422

Hope it's clear.


THanks for the reference Bunuel. So I am guessing its best to write out the factors in these types of questions? I didn't quite understand your methodology of replacing 2, 2^2, and 5 by 7 etc. I understand how it is applied here but if this were a different question I'd be lost. I think the key is just to mulitply the different factors to arrive at numbers larger than 77.

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Re: factors [#permalink]

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New post 12 Nov 2010, 19:05
So qn=700=2^2*5^2*7 has 18 factors, on the gmat how do you quickly determine 5 out of the 18 factors are greater than 77 within 2-2.5 mins? i got kind of lost with the substitution method also.

Kudos [?]: 96 [0], given: 57

Re: factors   [#permalink] 12 Nov 2010, 19:05
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