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131524248137 is a prime number
BaseRepresentation
bin111101001111101110…
…1001100001001001001
3110120111011002001222112
41322133131030021021
54123330201420022
6140231005032105
712334202443322
oct1723735141111
9416434061875
10131524248137
1150863011337
12215a7255635
13c5309069c9
146519ac1449
15364ba9c8e2
hex1e9f74c249

131524248137 has 2 divisors, whose sum is σ = 131524248138. Its totient is φ = 131524248136.

The previous prime is 131524248113. The next prime is 131524248157. The reversal of 131524248137 is 731842425131.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 128209828096 + 3314420041 = 358064^2 + 57571^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-131524248137 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 131524248094 and 131524248103.

It is not a weakly prime, because it can be changed into another prime (131524248157) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65762124068 + 65762124069.

It is an arithmetic number, because the mean of its divisors is an integer number (65762124069).

Almost surely, 2131524248137 is an apocalyptic number.

It is an amenable number.

131524248137 is a deficient number, since it is larger than the sum of its proper divisors (1).

131524248137 is an equidigital number, since it uses as much as digits as its factorization.

131524248137 is an evil number, because the sum of its binary digits is even.

The product of its digits is 161280, while the sum is 41.

The spelling of 131524248137 in words is "one hundred thirty-one billion, five hundred twenty-four million, two hundred forty-eight thousand, one hundred thirty-seven".