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find number of perfect squares which are also factors of n, a, b, and c are prime >2.

hmmm........... looks like you prepared/made this question. your question is little unclear. i am assuming it as "find the perfect squares that are also factors of n, a, b, and c and assuming that n, a, b, and c are different primes and are greater than 2".

a^4 can be selected in two ways i.e. a^2 and a^4 (assuming a^2 and a^2 in a^4 are same) = 2 ways
b^3 can be selected in 1 way i.e. b^2= 1 way
c^7 can be selected in 4 ways i.e. c^2, c^4 and c^6= 3 ways

they out to be: a^2.b^2.c^2 or a^4.b^2.c^4 or a^2.b^2.c^6..and so on.... i dont think we can count a^2.c^2.b^1 as a perfect square factor?...if we consider a^2*1 as a factor than this is a perfect square..b^2*1 is a perfect square factor...if we have a^2.b^2*1 also a factor etc...then 11 makes sense...

A number n( positive integer) when factorised can be written as a^4*b^3*c^7.find number of perfect squares which are factors of n.a,b,c are prime >2.

n = a^4 * b^3 * c^7

a^4 has 4/2=2 perfect squares b^3 has 3/2=1 perfect squares c^7 has 7/2=3 perfect squares

a^4*b^3 has min(4,3)/2 = 3/2 = 1 perfect square b^3*c^7 has min(3,7)/2 = 3/2 = 1 perfect square c^7*a^4 has min(7,4)/2 = 4/2 = 2 perfect squares

a^4*b^3*c^7 has min(4,3,7)/2 = 3/2=1 perfect square

total: 11

OH MY GOD! Where have I come?????? How can you guys diligently solve such problems?? I cant even comprehend what such a problem means many a times!
_________________

rahul..duttsit approach is very clear...if you cant follow it...try strenghthning your number properties skills...get your self a MGMAT number properties book...

basically the rule. is.prime^even is always a perfect square so we know that that N has prime factors a,b and c...so now all even powers of these factors would constitute a perfect square..

Hmmm let's see if this thinking is wrong.
For a we have 3 possibilities, a^0, a^2, a^4
For b we have 2 possibilities, b^0, b^2
For c we have 4 possibilities, c^0, c^2, c^4, c^6

So we would have 3*2*4=24 possibilities. The smallest of the perfect square factors would be 1, and the largest would be a^4b^2c^6.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Hmmm let's see if this thinking is wrong. For a we have 3 possibilities, a^0, a^2, a^4 For b we have 2 possibilities, b^0, b^2 For c we have 4 possibilities, c^0, c^2, c^4, c^6

So we would have 3*2*4=24 possibilities. The smallest of the perfect square factors would be 1, and the largest would be a^4b^2c^6.

a, b, and c, all, are prime and >2.

laxieqv wrote:

A number n( positive integer) when factorised can be written as a^4*b^3*c^7.find number of perfect squares which are factors of n.a,b,c are prime >2.

Last edited by HIMALAYA on 20 Nov 2005, 21:36, edited 1 time in total.

Well I understood he wants factors of n. And he says that a, b, c are primes that are greater than 2. Did I miss something? :hmm:

i think you are ok. i have been reading . (full stop) as , (coma).

laxieqv wrote:

A number n( positive integer) when factorised can be written as a^4*b^3*c^7. find number of perfect squares which are factors of n. a,b,c are prime >2.

different possible square factors:
a^2,a^4,b^2,c^2,c^4,c^6

Now for combinations of the above you can select one or less from each group therefore from the a group you can select none, a^2 or a^4 - 3 ways. from the b group you can select none or b^2 - 2 ways and from the c group you can select none, c^2, c^4 or c^6 - 4 ways.

Therefore total number of perfect squares that can be formed from the above are = 3*2*4 = 24 from which we will have to subtract one since one possibility was selecting none from each group. Therefore answer = 23.

Therefore total number of perfect squares that can be formed from the above are = 3*2*4 = 24 from which we will have to subtract one since one possibility was selecting none from each group. Therefore answer = 23.

does the bold part happen? ..i don't think so. Or do you mean the case of a^0 * b^0 * c^0 (= 1)? If so, this case should still be counted coz 1 is considered a perfect square. Btw, we all overlooked the case of a^0, b^0 and c^0 , which was brightly pointed out by sis Honghu