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131044314503 is a prime number
BaseRepresentation
bin111101000001011011…
…0011000110110000111
3110112020200222220210202
41322002312120312013
54121334321031003
6140111222241115
712316254203024
oct1720266306607
9415220886722
10131044314503
1150637110073
122149258619b
13c4853592cb
1464b207054b
15361e89a088
hex1e82d98d87

131044314503 has 2 divisors, whose sum is σ = 131044314504. Its totient is φ = 131044314502.

The previous prime is 131044314463. The next prime is 131044314587. The reversal of 131044314503 is 305413440131.

It is a happy number.

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 131044314503 - 216 = 131044248967 is a prime.

It is a super-2 number, since 2×1310443145032 (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 (131044314703) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65522157251 + 65522157252.

It is an arithmetic number, because the mean of its divisors is an integer number (65522157252).

Almost surely, 2131044314503 is an apocalyptic number.

131044314503 is a deficient number, since it is larger than the sum of its proper divisors (1).

131044314503 is an equidigital number, since it uses as much as digits as its factorization.

131044314503 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 8640, while the sum is 29.

Adding to 131044314503 its reverse (305413440131), we get a palindrome (436457754634).

The spelling of 131044314503 in words is "one hundred thirty-one billion, forty-four million, three hundred fourteen thousand, five hundred three".