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426151034051 is a prime number
BaseRepresentation
bin1100011001110001001…
…01001011110011000011
31111201222012100112200112
412030320211023303003
523440224131042201
6523434315341535
742534264010541
oct6147045136303
91451865315615
10426151034051
11154803170336
126a7111518ab
13312554834a2
14168a9370d91
15b1426830bb
hex633894bcc3

426151034051 has 2 divisors, whose sum is σ = 426151034052. Its totient is φ = 426151034050.

The previous prime is 426151034039. The next prime is 426151034053. The reversal of 426151034051 is 150430151624.

It is a strong prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-426151034051 is a prime.

It is a super-2 number, since 2×4261510340512 (a number of 24 digits) contains 22 as substring.

It is a Sophie Germain prime.

Together with 426151034053, it forms a pair of twin primes.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 426151033999 and 426151034017.

It is not a weakly prime, because it can be changed into another prime (426151034053) 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, 213075517025 + 213075517026.

It is an arithmetic number, because the mean of its divisors is an integer number (213075517026).

Almost surely, 2426151034051 is an apocalyptic number.

426151034051 is a deficient number, since it is larger than the sum of its proper divisors (1).

426151034051 is an equidigital number, since it uses as much as digits as its factorization.

426151034051 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 14400, while the sum is 32.

Adding to 426151034051 its reverse (150430151624), we get a palindrome (576581185675).

The spelling of 426151034051 in words is "four hundred twenty-six billion, one hundred fifty-one million, thirty-four thousand, fifty-one".