+0

# perfect squares

+3
458
3
+113

How many different perfect squares are factors of 20072007 ?

A) 2016032

B) 2016031

C) 2016030

D) 2016029

E) 2016028

Dec 9, 2017

#1
0

We can factor the base of 2007 as follows:

2007 =3^2 x 223

Now we can raise the above 2 factors to the power of 2007:

(3^2)^2007  x  223^2007

3^4014  x 223^2007. So from the 4014 exponent, we have 4014/2 =2007 Perfect Squares. And from the 2007 exponent, we have: 2007 -1 / 2 =1,003 Perfect Squares. Then we multiply them together, we get: 2007 x 1003=2,013,021 Perfect Squares. To this will add the sum of the same number of perfect squares, i.e., 2007 + 1003 =3010 Perfect Squares.

So the final total =[1,003 x 2,007] + 3,010 =2,016,031 in total.

And that is my attempt !!!.

P.S. I neglected to take "1" as a perfect square !!. Therefore, if we add 1 to the above total, we have: 2,016,031 + 1 =2,016,032 !!.

Dec 9, 2017
edited by Guest  Dec 9, 2017
edited by Guest  Dec 9, 2017
edited by Guest  Dec 9, 2017
#2
0

Look at the problem with this simple example:
2^4 x 3^4=2 "2s" and 2 "3s" =2 x 2 =4 squares. But the product of 2 x 3=6^4, which gives us another 4 squares for a total of 2 x 2 + 4 + 1=9 Perfect squares. So that if we find ALL the divisors of 2^4 x 3^4 =1,296.
ALL divisors of 1,296 =1 | 2 | 3 | 4 | 6 | 8 | 9 | 12 | 16 | 18 | 24 | 27 | 36 | 48 | 54 | 72 | 81 | 108 | 144 | 162 | 216 | 324 | 432 | 648 | 1296 (25 divisors) =9 Perfect squares.

Dec 9, 2017
#3
0

Prime factorize 2007 to get 3^2 * 223
This means that there are already 2007 squares in 2007^2007, due to the 3^2.
Divide 2007 by 2 to get the number of perfect squares for the 223, the second part of the prime factorization.
2007/2=1003
After this, you multiply 2007 by 1003, since the prime factorization is 3^2 multiplied by 223.
2007*1003=2013021
Then, you add the original 2007, since it it is 2007^2007, so there are 2007 perfect squares right there already for the 3^2 and add 1003 for the perfect squares from 223.
Add 1 also, because 1 is technically considered to be a perfect square (1*1=1)

2013021+2007+1003+1=2016032

Therefore, there are 2016032 factors in 2007^2007 that are perfect squares

Yahoo answers will always be better!

Dec 10, 2017