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

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

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.

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.

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

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.

Cheers!!!

Please repost your question. The language is not clear however one thing that I can suggest is "It's forbidden to cancel the commonn factors between Numerator and denominator for Remainder and factors questions"

Cancelling common factors between numerator and denominator changes the numbers

e.g.

when 3 is divided by 2, the remainder is 1

But, when 6 is divided by 4, the remainder is 2

and, when 9 is divided by 6, the remainder is 3

whereas (3/2) = (6/4) = (9/6)

We can create the equation:

777/N = Q + 77/N

When N = 700, the remainder is 77.

Thus, all factors of 700 that are greater than 77, will also leave a remainder of 77 when divided into 777.

Breaking 700 into primes, we have:

700 = 100 x 7 = 2^2 x 5^2 x 7^1

So factors of 700 that are greater than 77 are:

100, 7 x 25 = 175, 7 x 25 x 2 = 350, 7 x 4 x 5 = 140, and lastly, 700.

So there are 5 possibilities of N.

Answer: D

777 = Nk + 77
700 = Nk
N if a factor of 700 = 2^2*5^2*7 and N>77

N = {100, 175, 140, 350, 700}

IMO D

777=N*k+77

Where k is the quotient and 77 is the remainder.

N* k = 700 = \(7 * 2^2 * 5^2\)

Value of N should be greater than 77 as the divisor should be always greater than remainder.

There are (1+1)*(2+1)*(2+1)= 2*3*3= 18 factors possible for N.
Out of which the values greater than 77 are

1. \(7 * 2^2 * 5^2\) = 700

2. \(7*2*5^2\) = 350

3. \(7 * 5^2 \)= 175

4. \( 7 * 2^2 * 5\) = 140

5. \(2^2*5^2\) = 100

N*K = 700 can be written as 700 x 1, 350 x 2 ,175 x 4, 140 x 5, 100 x 7

Therefore, we can conclude that there are 5 possible values of N.

Option D is the answer.

Thanks,
Clifin J Francis,
GMAT SME

If we understand our factors well, we don't need to find out the large factors of 700.

As shown in multiple comments above, when 777 divided by N leaves remainder 77, it means N fully divides 700.

\(700 = 2^2 * 5^2 * 7\)
We need all factors greater than 77. We know that factors appear in pairs and that 700 = 10 * 70.
So we are looking for factors less than 10 since their pair will be greater than 77.
1, 2, 4, 5 and 7 are the 5 factors less than 10 and their pair will be greater than 77. We don't need to find the pair. We know that there are 5 such factors.

Answer (D)

Check this post: https://anaprep.com/number-properties-f ... -a-number/