What is the lowest number with 36 factors?

Answer 1

1260.

The number of factors is given by (a+1)*(b+1)... where N=x^a*y^b... is the prime factorization of N.

Note that 36=2^2*3^2, so (a+1)*(b+1)... has to equal that.

So, we know we have to include 2, 2, 3, and 3 in the equation (it can be separate or together, like (1+1) or (3+1), but it needs to come to that)

We can have a=2, b=2, c=1, d=1. (2+1)*(2+1)*(1+1)*(1+1)=3*3*2*2=36 This makes for 36 factors. With our exponents chosen, it would make the smallest number if we use the smallest primes for the largest exponents. 1260 = 2^2*3^2*5*7.

Or we could try a=3, b=2, c=2. (3+1)*(2+1)*(2+1)=4*3*3=36. Again, use smallest primes for biggest exponents. 1800 = 2^3*3 ^ 2*5 ^ 2

When you keep putting more of the numbers together, to the extremes of a=35 (35+1=36) or a=8, b=3 ((8+1)(3+1)=9*4=36), you get 2^35=34359738368 or 2^8*3^3=6912. I think our first guess was the best: 1260