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When the number 777 is divided by the integer N, the remaind

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

A. 2
B. 3
C. 4
D. 5
E. 6
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Last edited by Engr2012 on 17 Jul 2015, 03:58, edited 1 time in total.
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Re: When the number 777 is divided by the integer N, the remaind [#permalink]

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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.

Answer: 5.
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Re: When the number 777 is divided by the integer N, the remaind [#permalink]

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New post 15 Sep 2010, 14:39
Can N be allowed to be negative? Does GMAT allow that possibility and if so, how does one answer this then?
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New post 15 Sep 2010, 14:42
mainhoon wrote:
Can N be allowed to be negative? Does GMAT allow that possibility and if so, how does one answer this then?


No, it cannot be the case, at least for GMAT. Every GMAT divisibility question will tell you in advance that any unknowns represent positive integers.

So don't worry about that.
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Re: When the number 777 is divided by the integer N, the remaind [#permalink]

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

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

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New post 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 \(700\).
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Re: When the number 777 is divided by the integer N, the remaind [#permalink]

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New post 22 May 2011, 23:25
AtifS wrote:

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 \(700\).[/quote]

whats the fastest way of finding factors of 700 :roll:
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Re: When the number 777 is divided by the integer N, the remaind [#permalink]

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New post 23 May 2011, 07:20
Just break 700 into prime factors, take 5 and 2 as there is a 0 at end. Take 7 also, as the number is 700.

700 = 2^2 * 5^2 * 7
Then the number of factors will be (2+1) * (2+1) * (1+1)

Please ask if you have any more query.

Also, you can refer to Math Book for more details on prime factorization and number of factors.
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Re: When the number 777 is divided by the integer N, the remaind [#permalink]

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New post 23 May 2011, 18:35
700 = 2^2 * 5^2 * 7
How to come quickly with divisors greater than 77 from above expression? It took almost two minutes to me.
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Re: When the number 777 is divided by the integer N, the remaind [#permalink]

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New post 23 May 2011, 21:49
good concept of number > 77 here.

777-77 = nQ
2^2 * 5^2 * 7 = 700

gives 100,140,175,350 and 700 .
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Re: When the number 777 is divided by the integer N, the remaind [#permalink]

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

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New post 17 Oct 2013, 18:38
mainhoon wrote:
When the number 777 is divided by the integer N, the remainder is 77. How many integer possibilities are there for N?


Hey there, are there at least some answer choices for this question?
Much appreciated

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When the number 777 is divided by the positive integer n, the [#permalink]

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New post 17 Jul 2015, 04:01
Bunuel wrote:
When the number 777 is divided by the positive integer n, the remainder is 77. How many integer possibilities are there for n?

A. 2
B. 3
C. 4
D. 5
E. 6

Kudos for a correct solution.


The trick with this question is to realise that only numbers >77 will leave a remainder of 77 when dividing 777.

Given: 777=np+77 where n >77 ---> \(np =700 = 2^2*5^2*7\)

Now only numbers above 77 that will be factors of 700 are 100, 140, 175, 350 and 700. Thus 5 (D) is the correct answer.
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Hi Engr2012.. 145 cannot be correct. Did u mean to type 140 instead.

Engr2012 wrote:
Bunuel wrote:
When the number 777 is divided by the positive integer n, the remainder is 77. How many integer possibilities are there for n?

A. 2
B. 3
C. 4
D. 5
E. 6

Kudos for a correct solution.


The trick with this question is to realise that only numbers >77 will leave a remainder of 77 when dividing 777.

Given: 777=np+77 where n >77 ---> \(np =700 = 2^2*5^2*7\)

Now only numbers above 77 that will be factors of 700 are 100, 145, 175, 350 and 700. Thus 5 (D) is the correct answer.

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Re: When the number 777 is divided by the positive integer n, the [#permalink]

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New post 17 Jul 2015, 06:22
shriramvelamuri wrote:
Hi Engr2012.. 145 cannot be correct. Did u mean to type 140 instead.

Engr2012 wrote:
Bunuel wrote:
When the number 777 is divided by the positive integer n, the remainder is 77. How many integer possibilities are there for n?

A. 2
B. 3
C. 4
D. 5
E. 6

Kudos for a correct solution.


The trick with this question is to realise that only numbers >77 will leave a remainder of 77 when dividing 777.

Given: 777=np+77 where n >77 ---> \(np =700 = 2^2*5^2*7\)

Now only numbers above 77 that will be factors of 700 are 100, 145, 175, 350 and 700. Thus 5 (D) is the correct answer.


Yes, I meant 140. It was a typo.
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Re: When the number 777 is divided by the positive integer n, the [#permalink]

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Bunuel wrote:
When the number 777 is divided by the positive integer n, the remainder is 77. How many integer possibilities are there for n?

A. 2
B. 3
C. 4
D. 5
E. 6

Kudos for a correct solution.


There are only 5 such numbers = 100, 140, 175, 350 and 700
Answer D
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Re: When the number 777 is divided by the positive integer n, the [#permalink]

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Bunuel wrote:
When the number 777 is divided by the positive integer n, the remainder is 77. How many integer possibilities are there for n?

A. 2
B. 3
C. 4
D. 5
E. 6

Kudos for a correct solution.


777 = xn+77
700 = xn

Two numbers multiplied together must equal 700. To get a remainder above 77, each number must be above 77.
Prime factorization of 700 = 2^2*5^2*7

The answer choice are 100, 140, 175, 350, 700.

D
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Re: When the number 777 is divided by the positive integer n, the   [#permalink] 17 Jul 2015, 09:31

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