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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 remaind  [#permalink]

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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

Originally posted by mainhoon on 15 Sep 2010, 14:04.
Last edited by ENGRTOMBA2018 on 17 Jul 2015, 03:58, edited 1 time in total.
Edited the question, added the OA and tags
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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, 14:20
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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.

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 29 Jun 2013, 12:09
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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 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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When the number 777 is divided by the integer N, the remaind  [#permalink]

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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
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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


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

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New post 17 Jul 2015, 04:01
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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.


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

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New post 17 Jul 2015, 08:41
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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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New post 17 Jul 2015, 09:31
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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.

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

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New post 19 Jul 2015, 12:52
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.


800score Official Solution:

If the remainder is 77, then n must logically be greater than 77. Also, there must be a positive integer q such that 777= nq + 77. i.e. nq = 700. Therefore, the factors of 700 greater than 77 comprise the possible values of n.

Instead of counting the factors of 700 that are greater than 77, let’s count the ones that are less than or equal to 700/77 (or about 9).
As 700 = 50 × 2 × 7, we can see that there are 5 factors of 700 that are less than or equal to 9: 1 , 2 , 4 , 5 , and 7.
Thus there are 5 possible values of n (i.e. factors of 700) greater than 77. They are 700, 350, 175, 140 and 100.

A simpler way to think about this is to figure out all the divisors of 700 bigger than 77. That gives you 700, 350, 175, 140 and 100.

The correct answer is D.
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When the number 777 is divided by the integer N, the remaind  [#permalink]

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New post 19 Jul 2015, 19:41
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.


OBSERVATION-1: When 777 is divided by the positive integer n, the remainder is 77 i.e. 777-77 = 700 MUST be divisible by the divisor

OBSERVATION-2: Since the Remainder is 77 therefore the divisor MUST BE greater than 77


700 can be written as product of two Integers as follows

1*700
2*350
4*175
5*140
7*100
10*70
14*50
20*35
25*28

Out of all the factors mentioned above the Numbers satisfying the above mentioned conditions and Observations are {700, 350, 175, 140, 100}

Hence, 5 Numbers

Answer: Option D
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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 Jul 2015, 20:54
Dividend = Quotient x Divisor + Remainder
777 = Q*N + 77
Q*N = 700 = 2^2 * 5^2 * 7

Total values 18
But N cannot be less than 77,
100, 140, 175, 350, 700 (5 values)

I calculated manually and it took more than 65 seconds to get these values.
Any shortcut methods appreciated, though I can't think of any possibility.
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When the number 777 is divided by the integer N, the remaind  [#permalink]

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New post 19 Jul 2015, 20:58
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sanket1991 wrote:
Dividend = Quotient x Divisor + Remainder
777 = Q*N + 77
Q*N = 700 = 2^2 * 5^2 * 7

Total values 18
But N cannot be less than 77,
100, 140, 175, 350, 700 (5 values)

I calculated manually and it took more than 65 seconds to get these values.
Any shortcut methods appreciated, though I can't think of any possibility.



1) Where thinking manually is not a bad idea, it's not a great idea either if you don't have a thought of how are you going to get all the factors without missing on any one. So you can note down the number as product of two numbers and then make sure that all the numbers that need to be taken into account have been taken into account

2) Attempting this question in 65 seconds is a good speed already so I don't think any method can get you to answer in time less than 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 02 Oct 2015, 12:01
please bunuel help me in this. 77/700 = 11/100 is 700 in such case, a possibility for n of the 5 possibility for n ?
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Re: When the number 777 is divided by the integer N, the remaind   [#permalink] 02 Oct 2015, 12:01

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