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Re: When the number 777 is divided by the integer N, the remaind [#permalink]

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15 Sep 2010, 14:20

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mainhoon wrote:

When the number 777 is divided by the integer N, the remainder is 77. How many integer possibilities are there for N?

I think it should me mentioned that \(n\) is a positive integer.

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.

Re: When the number 777 is divided by the integer N, the remaind [#permalink]

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15 Sep 2010, 14:49

mainhoon wrote:

Can N be allowed to be negative? Does GMAT allow that possibility and if so, how does one answer this then?

the problem with negative numbers is that there is no unique definition of remainder

the only condition is that abs(remainder)<abs(divisor)

But even if we follow that, it is enough to tell us that the possible divisors is just double. All the positive ones listed above as well as -1*those numbers hence, 10 _________________

Re: When the number 777 is divided by the integer N, the remaind [#permalink]

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18 Oct 2010, 11:29

Bunuel wrote:

mainhoon wrote:

When the number 777 is divided by the integer N, the remainder is 77. How many integer possibilities are there for N?

I think it should me mentioned that \(n\) is a positive integer.

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.

Answer: 5.

I couldn't understand how we arrived at 100,140,175,350 and 700... is it something that we did manually or erupted out of the calculation given here _________________

Re: When the number 777 is divided by the integer N, the remaind [#permalink]

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19 Oct 2010, 02:30

hirendhanak wrote:

Bunuel wrote:

mainhoon wrote:

When the number 777 is divided by the integer N, the remainder is 77. How many integer possibilities are there for N?

I think it should me mentioned that \(n\) is a positive integer.

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.

Answer: 5.

I couldn't understand how we arrived at 100,140,175,350 and 700... is it something that we did manually or erupted out of the calculation given here

by multiplying the factors of \(700\) with each other and selecting only those numbers which result in \(\geq 77\) till \(700\). _________________

"Don't be afraid of the space between your dreams and reality. If you can dream it, you can make it so." Target=780 http://challengemba.blogspot.com Kudos??

Re: When the number 777 is divided by the integer N, the remaind [#permalink]

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29 Jun 2013, 12:09

4

This post received KUDOS

hirendhanak wrote:

Bunuel wrote:

mainhoon wrote:

When the number 777 is divided by the integer N, the remainder is 77. How many integer possibilities are there for N?

I think it should me mentioned that \(n\) is a positive integer.

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.

Answer: 5.

I couldn't understand how we arrived at 100,140,175,350 and 700... is it something that we did manually or erupted out of the calculation given here

One way to think about this that might help is the following:

You know that the feasible factors of 700 must be in a range above 77. Thus start breaking down 700 "from the top":

700 x 1 350 x 2 175 x 4 140 x 5 100 x 7

The next one, 70 x 10, is already out of range. That gives you 5 factors.

Re: When the number 777 is divided by the integer N, the remaind [#permalink]

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13 Nov 2014, 16:08

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