Find the smallest positive integer N

wvgewgeg

New member
Find the smallest positive integer N that satisfies all of the following conditions:
• N is a square.
• N is a cube.
• N is an odd number.
• N is divisible by twelve prime numbers.
How many digits does this number N have?

Subhotosh Khan

Super Moderator
Staff member
Find the smallest positive integer N that satisfies all of the following conditions:
• N is a square.
• N is a cube.
• N is an odd number.
• N is divisible by twelve prime numbers.
How many digits does this number N have?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

pka

Elite Member
Find the smallest positive integer N that satisfies all of the following conditions:
1. • N is a square.
2. • N is a cube.
3. • N is an odd number.
4. • N is divisible by twelve prime numbers.
How many digits does this number N have?
The number 1 has all but property #4. There is a start.

wvgewgeg

New member
The trouble is the divisibility by the first 12 prime numbers,
so it must be a multiple of 2*3*5*7*11*13*17*19*23*29*31*37

To be odd it must look like 2K+1

to be a square it must look like (2K+1)^2, and it must also be a cube
it must contain (2K+1)^6

so, it must have the form:
2*3*5*7*11*13*17*19*23*29*31*37(2K+1)^6
when K = 0, we get
2*3*5*7*11*13*17*19*23*29*31*37(1)^6
= 7.420738135... x 10^12
which would be 13 digits long

Please correct me if I am wrong!

I am really trying, please if anybody can solve this.

wvgewgeg

New member
The trouble is the divisibility by the first 12 prime numbers,
so it must be a multiple of 2*3*5*7*11*13*17*19*23*29*31*37

To be odd it must look like 2K+1

to be a square it must look like (2K+1)^2, and it must also be a cube
it must contain (2K+1)^6

so, it must have the form:
2*3*5*7*11*13*17*19*23*29*31*37(2K+1)^6
when K = 0, we get
2*3*5*7*11*13*17*19*23*29*31*37(1)^6
= 7.420738135... x 10^12
which would be 13 digits long

Please correct me if I am wrong!

I am really trying, please if anybody can solve this.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

pka

Elite Member
The trouble is the divisibility by the first 12 prime numbers,
so it must be a multiple of 2*3*5*7*11*13*17*19*23*29*31*37
I am really trying, please if anybody can solve this.
You want the first twelve odd primes. The product cannot include 2 if it is to be odd.
The number [imath]2374061173201265625[/imath] is the smallest positive integer that is:
1. odd
2. a square
3. a cube
4. is a multiple of the first four odd primes.
Now you wvgewgeg must complete the task and post a result.

Jomo

Elite Member
As pka stated, any integer multiple of 2 as in 2*(3*5*7*11*13*17*19*23*29*31*37) must be even.
I just needed to repeat that point!

Given a prime p, what can you do to this number to make it a perfect square and a perfect cube. That is, what can you have to multiply p by to get a perfect square and perfect cube?

Once you figure that out, you'll know the answer.

JeffM

Elite Member
Your first thought of testing the number one is good. Zero and one are numbers with interesting properties. You rejected it. Good work.

Your second thought of using the 12 smallest primes fits the thought of “smallest integer,” but does not fit the thought of “odd.” Why would you not reject 12 smallest primes out of hand. That leads to the product of the 12 smallest odd primes. Call that number x.

Now the question is whether you told us the problem correctly.

Is x a non-negative real number? Does it have a square root? Does it have a cube root? So is x the answer?

Is x a perfect square? Is x a perfect cube? So is x the answer?

Last edited:

pka

Elite Member
Your first thought of testing the number one is good. Zero and one are numbers with interesting properties. You rejected it. Good work.

Your second thought of using the 12 smallest primes fits the thought of “smallest integer,” but does not fit the thought of “odd.” Why would you not reject 12 smallest primes out of hand. That leads to the product of the 12 smallest odd primes. Call that number x.

Now the question is whether you told us the problem correctly.

Is x a non-negative real number? Does it have a square root? Does it have a cube root? So is x the answer?

Is x a perfect square? Is x a perfect cube? So is x the answer?
The positive odd integer [imath]\large N=3\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23\cdot 29\cdot 31\cdot 37\cdot 41=152125131763605[/imath]
is the smallest positive integer that is odd and is divisible by the first twelve odd primes.
Now [imath]N^6[/imath] is the smallest positive integer that is odd, is divisible by the first twelve odd primes, and is both a square & a cube.