[khanna_sumit]

Originally Posted by amar_kashyap

problem#3
A number N has 3 prime factors and 27 factors which are perfect cubes. If 125 of the factors of N are perfect squares , how many factors does N have...

A.648 B. 729 C. 900 D. 1000.

getting the answer as 729....

[amar_kashyap]

Originally Posted by khanna_sumit

getting the answer as 729....

How did u do it..??. I think ur answer is right..

[the_CONSTANTINE]

well..

Beats me how to do this question by conventional methods....
But if we closely look upon the options and the figures given, 729 becomes the natural answer......Well..Why??

Look at all the numbers....When the number is having such (so many)factors, it is obvious that the prime number shall be either 2,3, 5, or 7....
Now, powers of all the others will become too unwieldly...Except 2 or 3...
In options, 729 seems to be a good one(I wud hav been confused if there had been 512)

I know...This is not the correct line of th8, but still.....
Cheers
Looking for an appropriate soln from Ur side....
(PS: Forgive me..I am pathetic at Quants...)

[varun nakra1]

Originally Posted by amar_kashyap

problem#3
A number N has 3 prime factors and 27 factors which are perfect cubes. If 125 of the factors of N are perfect squares , how many factors does N have...

A.648 B. 729 C. 900 D. 1000.

Let N= a^p * b^q * c^r where a,b and c are all prime..
Now all the factors of N which are perfect squares will be any combination frm the following sets
{a^2 ,a^4 ,a^6,.......,a^m1}
{b^2 ,b^4 ,b^6,.......,a^m2}
{c^2 ,c^4 ,c^6,........,a^m3} such that m1<p ;m2<q and m3<r and m1,m2,m3 are all even.
So no. of factors that can be perfect squares =(1+m1/2)(1+m2/2)(1+m3/2)

Similarly no of factors that can be perfect cubes=(1+k1/3)(1+k2/3)(1+k3/3) such that k1<p ;k2<q ;k3<r and k1,k2,k3 are all divisible by 3.

So we have (1+m1/2)(1+m2/2)(1+m3/2) = 125 =5.5.5
and (1+k1/3)(1+k2/3)(1+k3/3) =27 =3.3.3
Then we can safely assume m1=m2=m3=8 and k1=k2=k3=6
Now if we can that N= a^8 * b^8 *c^8 we can have 9^3=729 factors
coz we cant say that any of p,q or r >8 then that would increase the number of perfect cubes(coz,the next power will be 9)so max of p,q and r is each of them =8 and the answer is 729 total factors.

P.S.-I am dead sure that there is better method of doing it.
P.S.S.- I may be wrong. So plz analyse it further and point out my mistakes

[Aarav]

Let p, q, r be 3 prime factors

=> sum of the divisors is (1+p+p^2 + ...)(1+q+q^2 + ...)(1+r+r^2+...)

Now, 125 factors are perfect squares => each of the above expression takes 5 terms which are perfect squares. E.g. first expression has terms 1, p^2, p^4, p^6, p^8
Why the information that it has 27 cubes is needed? B'coz if we say it has 125 squares but
64 cubes then we have 1000 factors as p^9, q^9, r^9 are also the factors.

[Anas]

Originally Posted by amar_kashyap

So here goes the first question: I am planning to move backward on the MB thread of previous year from 50th page to 1st page.

problem #1
What is tens digit in 32^32?

by another method answer is coming 5, it is as follows:
by making cycle of digits which come in ten's place by successively squaring 2 we get
2,4,8,1,3 as first five digits which do not repeat.
But then 6,2,5,1,2,4,9,8. This eight digits repeat therafter in ten's place.
So 32^32= 2^160

2^157= has 8 in tens place and now from cycle of eight digits mention above
2^158= 6
2^159= 2
2^160= 5 = answer

let me know wether this approach is right and also d ans!!!!!!!

[Anas]

the answer is coming 5 by another method which is as follows:
by taking succesive square of 2 we can make a cycle of digits in ten's place which is not found in first 5 digits.

first five digits are 2,4,8,1,3.
But then 6,2,5,1,2,4,9,8. these eight digits are repeated thereafter

So 32^32= 2^160
In 2^157 the ten's digit has 8.
and from cycle given above 2^158= has 6
2^159= has 2
2^160= has 5 in tens place = answer!!!!!
:grab:
Let me know wether this approach is right or wrong

[amar_kashyap]

Originally Posted by Anas

the answer is coming 5 by another method which is as follows:
by taking succesive square of 2 we can make a cycle of digits in ten's place which is not found in first 5 digits.

first five digits are 2,4,8,1,3.
But then 6,2,5,1,2,4,9,8. these eight digits are repeated thereafter

So 32^32= 2^160
In 2^157 the ten's digit has 8.
and from cycle given above 2^158= has 6
2^159= has 2
2^160= has 5 in tens place = answer!!!!!
:grab:
Let me know wether this approach is right or wrong

Bhai mere ..kuch samjha nahi..
And you have double posted.Plz have alittle patience while posting..

[IdiotR]

Originally Posted by amar_kashyap

Bhai mere ..kuch samjha nahi..
And you have double posted.Plz have alittle patience while posting..

same here

02 04 08 16 32 64 128 256 512 1024 2048 4096 8192 16384
# # # # # # #
kahan repeat ho raha hein yaar ?
What i know is tht if we divide successive powers of 2 by 100 thr will b a number which will leave remainder as 1 when divided by 100 and any multiple of that power will also leave 1 as remainder,thtz how we solve remainder probs...

[khanna_sumit]

Originally Posted by varun nakra1

Another method is write 32^32 =(30+2)^32 Now opening the binomial expression gives us the last two terms as 30*32*2^31 whch gives (....80) as a number and the last term as 2^32..Now 2^32 =[(1024)^3]*4..then 24*24 gives 76 as the last two digits then 76*24 again gives us 24 and finally 24*4 gives 96 as the last two digits (...96)..Thus the binonial expression gives us all zeroes at the unit's place and a 6 at the unit's place coz of the last term..and it gives all zeroes at the tenth place and one 9 and one 8 at the tenth place..i.e. .........00+......00+........+.......80+........96 =............76 as the answer

But i still feel modulo by 100 is a better method

modulo method.........

remainder when 2^160 is divided by 100?
cancelling common factors the problem reduces to finding the remainder when the number 2^158 is divided by 25......
euler's number for 25 is = 20 hence 2^20 divided by 25 gives a remainder 1.
using this result the problem reduces to finding the remainder when 2^18 is divided by 25. lets say this remainder is r. then 2^18 = 25q + r.
we know that 2^20 = 25p + 1
that implies 2^18 (mod 25) * 2^2 (mod 25) = 1 (mod 25)
or 4r = 25p + 1 which is satisfied at p = -1
hence r = -6 = 25 - 6 = 19...
a common factor 4 was cancelled initially therefore 19*4 = 76 is the answer..
It is difficult to explain how to use this method but believe me very easy to apply for any general problem once u understand it.

regards.