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131131952557 is a prime number
BaseRepresentation
bin111101000100000010…
…0101100110110101101
3110112110210220101121121
41322020010230312231
54122024234440212
6140124032500541
712321400130404
oct1721004546655
9415423811547
10131131952557
1150681628911
12214b79aa751
13c49a562148
1464bd96463b
15362740bd07
hex1e8812cdad

131131952557 has 2 divisors, whose sum is σ = 131131952558. Its totient is φ = 131131952556.

The previous prime is 131131952551. The next prime is 131131952563. The reversal of 131131952557 is 755259131131.

It is a balanced prime because it is at equal distance from previous prime (131131952551) and next prime (131131952563).

It can be written as a sum of positive squares in only one way, i.e., 113123922921 + 18008029636 = 336339^2 + 134194^2 .

It is a cyclic number.

It is not a de Polignac number, because 131131952557 - 27 = 131131952429 is a prime.

It is a super-2 number, since 2×1311319525572 (a number of 23 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (131131952551) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65565976278 + 65565976279.

It is an arithmetic number, because the mean of its divisors is an integer number (65565976279).

Almost surely, 2131131952557 is an apocalyptic number.

It is an amenable number.

131131952557 is a deficient number, since it is larger than the sum of its proper divisors (1).

131131952557 is an equidigital number, since it uses as much as digits as its factorization.

131131952557 is an evil number, because the sum of its binary digits is even.

The product of its digits is 141750, while the sum is 43.

The spelling of 131131952557 in words is "one hundred thirty-one billion, one hundred thirty-one million, nine hundred fifty-two thousand, five hundred fifty-seven".