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131024431457 is a prime number
BaseRepresentation
bin111101000000110101…
…0100010100101100001
3110112012022120202101122
41322001222202211201
54121314223301312
6140105232142025
712315623201663
oct1720152424541
9415168522348
10131024431457
115062696a6a7
1221487997915
13c4811b71aa
1464ad576533
15361cc6db72
hex1e81aa2961

131024431457 has 2 divisors, whose sum is σ = 131024431458. Its totient is φ = 131024431456.

The previous prime is 131024431441. The next prime is 131024431469. The reversal of 131024431457 is 754134420131.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 68576420641 + 62448010816 = 261871^2 + 249896^2 .

It is a cyclic number.

It is not a de Polignac number, because 131024431457 - 24 = 131024431441 is a prime.

It is a super-2 number, since 2×1310244314572 (a number of 23 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (131024431427) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65512215728 + 65512215729.

It is an arithmetic number, because the mean of its divisors is an integer number (65512215729).

Almost surely, 2131024431457 is an apocalyptic number.

It is an amenable number.

131024431457 is a deficient number, since it is larger than the sum of its proper divisors (1).

131024431457 is an equidigital number, since it uses as much as digits as its factorization.

131024431457 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 40320, while the sum is 35.

Adding to 131024431457 its reverse (754134420131), we get a palindrome (885158851588).

The spelling of 131024431457 in words is "one hundred thirty-one billion, twenty-four million, four hundred thirty-one thousand, four hundred fifty-seven".