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# Number Theory Help

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Of all the 900,000 6-digit integers, how many of them are both perfect squares and perfect cubes for the same number? Thanks for help.

Dec 13, 2019

#3
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If I understand the question:
A number is perfect square and a perfect cube if it follows this rule: If its prime factor(s) is a multiple of 6. So, you would try the first few small primes raised to power of: 6, 12, 18.......etc:

2^6 = 64 too small. 2^12 =4096 still too small. 2^18 =262144 - just right! so that one number.
3^6 =729 too small. 3^12 =531441 - just right. So that is your 2nd number.
5^6 =15625 - too small. 5^12 =244,140,625 way too large.
7^6 =117649 - just right. So, that is your 3rd number.
11^6 =1,771,561 - too large.
And that is it. You have ONLY 3 such numbers:

Number        Sq.root       Cubic root
117649           343             49
262144           512             64
531441           729             81

Dec 13, 2019

#1
+21725
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How can they be perfect squres and pefect cubes of the SAME number?????

Do you mean perfect squares and perfect cubes of DIFFERENT numbers?...or something else?

Dec 13, 2019
#2
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Sorry EP: I mean the same 6-digit number is a perfect square AND a perfect cube of  2 different numbers such as 4096 which is a perfect square of 64 and a perfect cube of 16.

Dec 13, 2019
#3
+1

If I understand the question:
A number is perfect square and a perfect cube if it follows this rule: If its prime factor(s) is a multiple of 6. So, you would try the first few small primes raised to power of: 6, 12, 18.......etc:

2^6 = 64 too small. 2^12 =4096 still too small. 2^18 =262144 - just right! so that one number.
3^6 =729 too small. 3^12 =531441 - just right. So that is your 2nd number.
5^6 =15625 - too small. 5^12 =244,140,625 way too large.
7^6 =117649 - just right. So, that is your 3rd number.
11^6 =1,771,561 - too large.
And that is it. You have ONLY 3 such numbers:

Number        Sq.root       Cubic root
117649           343             49
262144           512             64
531441           729             81

Guest Dec 13, 2019
#4
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Thank you.

Dec 14, 2019