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Factorization

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Factorization

by mitaliisrani » Sun May 23, 2010 12:09 am
252 toys of same type are to be given to 'c' number of children, such that 1<=c<=252. Each child should get an equal number of toys. How many values can 'c' take ?

A) 16
B) 17
C) 18
D) 19
E) 20

Well my approach was to factorize 252= 2*2*3*3*7

Then i realised that c can take multiple values such as 2,3,7,6,14 but then i got lost in the process

Plz help


OA:c
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by liferocks » Sun May 23, 2010 12:19 am
252= 2*2*3*3*7
or 252= 2^2*3^2*7

So total number of factors of 252,including 1 and 252 is (2+1)*(2+1)*(1+1)=18

hence number of values of C can be 18
Ans option C
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by frank1 » Sun May 23, 2010 4:09 am
My gmat approach
(One should always bear that gmat approach and normal approach are completely different things)

it should be divided among equal person
so what ever gives 0 remainder is answer here
and while guessing always start from C (lol....hope gmac people are not watching ...)
so 252/18 .....wow....no remainder ...i work as 180+90-18.....i dont remember table of 18
so done...C
i dont bother to check others....
if there are other such option with remainder 0 then question is wrong....

but question writer are cleaver....
they will say ...max value c can take....
then it will require some more work...
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edited for typo

by gmatjedi » Sun May 23, 2010 6:31 am
my approach:

goal is to calculate the number of combinations of numbers that will divide 252 into an integer

step 1: factor 252 into primes = 2^2*3^2*7
step 2: number of combinations of such numbers= 3 choices for 2 (zero 2s, one 2, or two 2s)*3 choices for 3 (zero 3s, one 3, or two 3s)*2 choices for 7 (zero 7s, or one seven)= 3*3*2=18
Last edited by gmatjedi on Sun May 23, 2010 8:21 am, edited 1 time in total.
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by gmatmachoman » Sun May 23, 2010 7:39 am
gmatjedi wrote:my approach:

goal is to calculate the number of combinations of numbers that will divide 252 into an integer

step 1: factor 252 into primes = 2^2*3^2*7
step 2: number of combinations of such numbers= 3 choices for 2 (zero 2s, one 2, or two 2s)*3 choices for 3 (zero 3s, one 3, or two 3s)*2 choices for 7 (zero 7s, or one seven)= 3*3*2=18
Corrected ur typo error!!( look for bold letters)
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by gmatjedi » Sun May 23, 2010 8:23 am
thanks gmatmachoman
fixed the typo
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by neha.patni » Sun May 23, 2010 9:34 pm
mitaliisrani wrote:252 toys of same type are to be given to 'c' number of children, such that 1<=c<=252. Each child should get an equal number of toys. How many values can 'c' take ?

A) 16
B) 17
C) 18
D) 19
E) 20

Well my approach was to factorize 252= 2*2*3*3*7

Then i realised that c can take multiple values such as 2,3,7,6,14 but then i got lost in the process

Plz help


OA:c
IMO C

I followed the reverse approach i.e. dividing 252 from the given options
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