Repeated prime factors
360 = 2^3 × 3^2 × 5. Its positive divisor count is (3 + 1)(2 + 1)(1 + 1) = 24.
Break a positive integer into its prime factors, group repeated factors as powers, and see how many positive divisors the number has. The result preserves the exact integer values throughout the calculation.
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Enter an integer from 2 through 1,000,000,000,000 using digits only.
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360 = 2^3 × 3^2 × 5. Its positive divisor count is (3 + 1)(2 + 1)(1 + 1) = 24.
169 = 13^2. Its three positive divisors are 1, 13, and 169.
In the prime-power product, the exponent counts how often the prime appears. For example, 2^3 stands for 2 × 2 × 2. A prime whose exponent is 1 appears once and is written without a power in the product.
If the factorization is p^a × q^b, a positive divisor can contain any power of p from 0 through a and any power of q from 0 through b. Multiplying the number of choices for each prime gives the divisor count. This count includes both 1 and the original number.
The calculator divides out every factor of 2 and 3, then tests candidates on either side of multiples of 6. Each time it finds a factor, it removes all copies. Any remaining value above 1 after the square-root stopping condition is itself prime.
One has an empty prime factorization. This tool requires at least 2 so every successful result contains a prime factor.
The product contains only that prime with exponent 1. Its positive divisor count is 2.
Yes. Parsing, division, and remainder checks use BigInt integers. No approximate numerical root finding is involved.
Use the absolute value here. The sign −1 is a unit rather than a prime factor and is outside this tool’s positive-input model.