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1300003100113 is a prime number
BaseRepresentation
bin10010111010101110001…
…110010001010111010001
311121021112110011221210011
4102322232032101113101
5132244401243200423
62433114014034521
7162631201012534
oct22725616212721
94537473157704
101300003100113
1146136794a373
1218bb484a8a41
1395788941523
1446cc5b54a1b
1523c39338b0d
hex12eae3915d1

1300003100113 has 2 divisors, whose sum is σ = 1300003100114. Its totient is φ = 1300003100112.

The previous prime is 1300003100107. The next prime is 1300003100117. The reversal of 1300003100113 is 3110013000031.

It is a happy number.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 706905963729 + 593097136384 = 840777^2 + 770128^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-1300003100113 is a prime.

It is a super-2 number, since 2×13000031001132 (a number of 25 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 1300003100093 and 1300003100102.

It is not a weakly prime, because it can be changed into another prime (1300003100117) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 650001550056 + 650001550057.

It is an arithmetic number, because the mean of its divisors is an integer number (650001550057).

Almost surely, 21300003100113 is an apocalyptic number.

It is an amenable number.

1300003100113 is a deficient number, since it is larger than the sum of its proper divisors (1).

1300003100113 is an equidigital number, since it uses as much as digits as its factorization.

1300003100113 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 27, while the sum is 13.

Adding to 1300003100113 its reverse (3110013000031), we get a palindrome (4410016100144).

The spelling of 1300003100113 in words is "one trillion, three hundred billion, three million, one hundred thousand, one hundred thirteen".