BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Number system

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 167
Joined: Fri Mar 09, 2012 8:35 pm
Thanked: 39 times
Followed by:3 members

Number system

by adthedaddy » Wed Oct 10, 2012 8:22 am
How many divisors of 25200 can be expressed in the form 4n + 3, where n is a whole number?
(a) 6
(b) 8
(c) 9
(d) None of these

OA: [spoiler](c)[/spoiler]
"Your time is limited, so don't waste it living someone else's life. Don't be trapped by dogma - which is living with the results of other people's thinking. Don't let the noise of others' opinions drown out your own inner voice. And most important, have the courage to follow your heart and intuition. They somehow already know what you truly want to become. Everything else is secondary" - Steve Jobs
Top
Source: — Problem Solving |
User avatar
Master | Next Rank: 500 Posts
Posts: 279
Joined: Mon Jun 25, 2012 10:56 pm
Thanked: 60 times
Followed by:10 members

by anuprajan5 » Wed Oct 10, 2012 9:39 am
HI,

The answer is C

My logic was to break down the number into its primes which will give us 2^5*3^2*5^3*7

The form of the expression is odd (4n+3 - Even + Odd)

I calculated the odd divisors using the odd primes ie: 3,5,7

You get a set of numbers. All one needs to do is ensure that it of the form 4n+3

The divisors are 3,7,15,35,63,75,175,375,875

Regards
Anup
Top
User avatar
Senior | Next Rank: 100 Posts
Posts: 53
Joined: Sat Oct 29, 2011 3:49 am
Thanked: 2 times
Followed by:1 members

by gmat6087 » Wed Oct 10, 2012 11:02 am
anuprajan5 wrote:HI,

The answer is C

My logic was to break down the number into its primes which will give us 2^5*3^2*5^3*7

The form of the expression is odd (4n+3 - Even + Odd)

I calculated the odd divisors using the odd primes ie: 3,5,7

You get a set of numbers. All one needs to do is ensure that it of the form 4n+3

The divisors are 3,7,15,35,63,75,175,375,875

Regards
Anup
Anup,
I too ended up in the same way finding all the divisors that satisfy the above condition, is there any easier way to solve this instead of this lengthy method

Regards,
Satya
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 167
Joined: Fri Mar 09, 2012 8:35 pm
Thanked: 39 times
Followed by:3 members

by adthedaddy » Thu Oct 11, 2012 9:55 am
anuprajan5 wrote:HI,

The answer is C

My logic was to break down the number into its primes which will give us 2^5*3^2*5^3*7

The form of the expression is odd (4n+3 - Even + Odd)

I calculated the odd divisors using the odd primes ie: 3,5,7

You get a set of numbers. All one needs to do is ensure that it of the form 4n+3

The divisors are 3,7,15,35,63,75,175,375,875

Regards
Anup
@ Anup:
I can see above that you've substitued values of "n" in 4n+3. Plz help me understand how did you derive the value of "n". For example, when we subsitute n=0, we get 3, n=1 gives 7, then you have not taken n=2 but you have taken n=3...etc...
Plz help.
"Your time is limited, so don't waste it living someone else's life. Don't be trapped by dogma - which is living with the results of other people's thinking. Don't let the noise of others' opinions drown out your own inner voice. And most important, have the courage to follow your heart and intuition. They somehow already know what you truly want to become. Everything else is secondary" - Steve Jobs
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 279
Joined: Mon Jun 25, 2012 10:56 pm
Thanked: 60 times
Followed by:10 members

by anuprajan5 » Thu Oct 11, 2012 10:33 pm
Hi,

I wouldn't state that the method used by me is perfect. But this is how I did it.

Since the expression is odd - we use the odd divisors to find a set of numbers. You will find 18 divisors (based on powers of 3,5 and 7)


1,3,5,7,9,15,21,25,35,45,63,75,105,175,225,315,525,1575

Looking at this list now, I have to retract my answer from before (i calculated the same as 252000 and not 25200 - luckily though I got the same answer). It seems there are 9 divisors in the form 4n+3

3,7,15,35,63,75,175,315,1575

One you have the set above its easy to determine which ones fit the expression because of divisibility rule. If you take each individual number and deduct 3 from the same - and if the last digits are divisible by 4, then it matches the expression.

Regards
Anup
Top
Post new topic Post Reply
• Page 1 of 1