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3907750500133 is a prime number
BaseRepresentation
bin111000110111011000000…
…101001001011100100101
3111211120120211222120002221
4320313120011021130211
51003011033112001013
612151110143032341
7552216462124654
oct70673005113445
914746524876087
103907750500133
1112772a4311a87
1253142189b6b1
13224664c3325c
14d71c979b19b
156b9b23d508d
hex38dd8149725

3907750500133 has 2 divisors, whose sum is σ = 3907750500134. Its totient is φ = 3907750500132.

The previous prime is 3907750500131. The next prime is 3907750500229. The reversal of 3907750500133 is 3310050577093.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 3741350142049 + 166400358084 = 1934257^2 + 407922^2 .

It is a cyclic number.

It is not a de Polignac number, because 3907750500133 - 21 = 3907750500131 is a prime.

Together with 3907750500131, it forms a pair of twin primes.

It is a self number, because there is not a number n which added to its sum of digits gives 3907750500133.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (3907750500131) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1953875250066 + 1953875250067.

It is an arithmetic number, because the mean of its divisors is an integer number (1953875250067).

Almost surely, 23907750500133 is an apocalyptic number.

It is an amenable number.

3907750500133 is a deficient number, since it is larger than the sum of its proper divisors (1).

3907750500133 is an equidigital number, since it uses as much as digits as its factorization.

3907750500133 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 297675, while the sum is 43.

The spelling of 3907750500133 in words is "three trillion, nine hundred seven billion, seven hundred fifty million, five hundred thousand, one hundred thirty-three".