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 Post subject: GMAT Number TheoryPosted: Wed Mar 02, 2011 6:12 am

Joined: Sun May 30, 2010 3:15 am
Posts: 424
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

(D) 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.
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Why do we count the factors that are less than or equal to 700/77 ?

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 Post subject: Re: math (t.2, q.14): number propertiesPosted: Wed Mar 02, 2011 8:41 am

Joined: Sun May 30, 2010 2:23 am
Posts: 498
Imagine, that we have a factor of 700, a, which is greater than 77. Then we know that:
a > 77
700 = ab, so a = 700/b.
Therefore 700/b > 77 or 700/77 > b

As you see, for each factor a, which is greater than 77, there is a corresponding factor b = 700/a, which is less than 700/77.

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Furthermore, it's a very good property to remember. It helps to analyze if an integer is a prime number or to find all the possible factors. For example:

How many distinct factors does 127 have?

It's easy to see that 11 < √127 < 12.
As we already know from the above reasoning, any factor a < √127 will yield a factor b > 127/√127 = √127. So we have to check divisibility by only 11 numbers and that will give us all the possible divisors.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11

1) 127 = 1 × 127 is obvious. So we have two factors here.
2) 127 is an odd number. So it's not divisible by any even number.

This leaves us:
3, 5, 7 , 9, 11

3) 127 is NOT divisible by 3, since 1 + 2 + 7 = 11 is NOT.
4) 127 is NOT divisible by 5, since it doesn't end in 0 or 5.

This leaves us:
7, 11

5) 127 = 121 + 6 = 11² + 6, so it's NOT divisible by 11
6) 127/7 = 16 × 7 + 1 (you can do it by long division method), so it's NOT divisible by 7.

As the result, we know that 127 is a prime number and has only two distinct factors: 1, 127.

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 Post subject: Re: math (t.2, q.14): number propertiesPosted: Wed Mar 02, 2011 5:31 pm

Joined: Wed Mar 02, 2011 5:30 pm
Posts: 2
Ok, so basically once you see that for each factor a, which is greater than 77, there is a corresponding factor b = 700/a, which is less than 700/77, I just need to check divisibility by all factors < 700/77 < 10 right?

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 Post subject: Re: math (t.2, q.14): number propertiesPosted: Fri Mar 04, 2011 12:43 pm

Joined: Sun May 30, 2010 2:23 am
Posts: 498
Yes, you just need to check divisibility of 700 by all integers less than 700/77 (which means less than or equal 9).

1, 2, 3, 4, 5, 6, 7, 8, 9.

1 is obvious 700 / 1 = 700
2 is obvious 700 / 2 = 350
700 is NOT divisible by 3, since 7 + 0 + 0 = 7 is NOT. Therefore 6 and 9 are NOT factors as well.
4 is a factor, since 00 is divisible by 4. 700 / 4 = 175
5 is a factor, since the unit digit is 0. 700 / 5 = 140
7 is obvious 700 / 7 = 100
8 is NOT a factor, since 700 / 4 = 175, which is odd.

Therefore all the possible factors that are greater than 77 are 700, 350, 175, 140, 100.

NOTE, that you do NOT have to calculate all those factors that are greater than 77. I just showed those for reference. It's enough to calculate factors less than 700/77: 1, 2, 4, 5, 7.

Try to feel these two key moments here (that can be applied to many other problems):
1. Every factor a has a pair factor b = 700/a.
Why is it so?
The definition of a divisor, 700 = ab directly implies this rule.
If 700 = a², then b = 700/a = a. Otherwise a and b are different.

2. Every factor less than 700/77 corresponds to a factor, which is greater than 77 and vice-versa.
Why is it so?
If a < 700/77, then
700 / b < 700/77
Multiple the both sides by 77b/700 (a positive number):
77 < b.

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 Post subject: Re: math (t.2, q.14): number propertiesPosted: Fri May 13, 2011 10:29 pm

Joined: Fri May 13, 2011 10:27 pm
Posts: 1
Why do we not consider other factors of 700 like 25 or 28?

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 Post subject: Re: math (t.2, q.14): number propertiesPosted: Wed May 18, 2011 5:35 pm

Joined: Sun May 30, 2010 2:23 am
Posts: 498
PRACHI SHARMA wrote:
Why do we not consider other factors of 700 like 25 or 28?

By definition the remainder (77) must NOT exceed a divisor.

When 777 is divided by 25, the remainder is 2. 777 = 25 × 31 + 2
When 777 is divided by 28, the remainder is 21. 777 = 28 × 27 + 21

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 Post subject: Re: math (t.2, q.14): number propertiesPosted: Fri May 20, 2011 10:33 am

Joined: Sun May 30, 2010 3:15 am
Posts: 424
I am not able to understand the explanation of this question.

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 Post subject: Re: math (t.2, q.14): number propertiesPosted: Fri May 20, 2011 10:35 am

Joined: Sun May 30, 2010 2:23 am
Posts: 498
questioner wrote:
I am not able to understand the explanation of this question.

Dear student, please specify which part of the explanation you do not understand.

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