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by shashank.ism » Tue Feb 09, 2010 7:12 am
How many integers exist such that not only are they multiples of 2008^2008 but also are factors of
2008^2020?

a 12
b 481
c 587
d 2008^(12)
e 637
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by ajith » Tue Feb 09, 2010 9:40 am
shashank.ism wrote:How many integers exist such that not only are they multiples of 2008^2008 but also are factors of
2008^2020?

a 12
b 481
c 587
d 2008^(12)
e 637
2008^2020/2008^2008 = 2008^12

Now the question is equivalent to how many factors are there for 2008^12

2008^12 = (8*251)^12
= 2^36 *251^12
No of factors = 37*13

481
B
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by thephoenix » Tue Feb 09, 2010 9:48 am
How many integers exist such that not only are they multiples of 2008^2008 but also are factors of
2008^2020?

multiples of 2008^2008 --->K=x* 2008^2008

factors of 2008^2020--->2008^2020*x/k---->2008^(2020-2008)=2008^12

2008^12=[(2^3)*251]^12=2^36 251^12

hence soln is 37*13=481
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by shashank.ism » Tue Feb 09, 2010 9:55 am
Nice approach to solve the problem..
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by harsh.champ » Tue Feb 09, 2010 1:42 pm
ajith wrote:
shashank.ism wrote:How many integers exist such that not only are they multiples of 2008^2008 but also are factors of
2008^2020?

a 12
b 481
c 587
d 2008^(12)
e 637
2008^2020/2008^2008 = 2008^12

Now the question is equivalent to how many factors are there for 2008^12

2008^12 = (8*251)^12
= 2^36 *251^12
No of factors = 37*13

481
B
Superb approach to solve the problem.
I had gone nuts figuring out how to solve the problem.

Can, you explain this step?

= 2^36 *251^12
No of factors = 37*13
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



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