Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Cube of Positive Integer.

Cube of Positive Numbers – AIME I, 2015

There is a prime number p such that 12p+1 is the cube of positive integer.Find p..

  • is 107
  • is 183
  • is 840
  • cannot be determined from the given information

Key Concepts


Theory of Equations

Number Theory

Check the Answer

But try the problem first…

Answer: is 183.

Suggested Reading

AIME, 2015, Question 3

Elementary Number Theory by David Burton

Try with Hints

First hint

\(a^{3}=12p+1\) implies that \(a^{3}-1=12p\) that is (a-1)(\(a^{2}\)+a+1)=12p

Second Hint

a is odd, a-1 even, \(a^{2} +a+1 odd implies a-1 multiple of 12 that is here =12 then a=12+1 =13

Final Step

\(a^{2}+a+1=p implies p= 169+13+1=183.

Subscribe to Cheenta at Youtube