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49998717504013 is a prime number
BaseRepresentation
bin10110101111001001110111…
…10010111100101000001101
320120000211102120220201101111
423113210323302330220031
523023134333140112023
6254201032125012021
713350200554402555
oct1327447362745015
9216024376821344
1049998717504013
1114a27378626953
1257361053b4011
1321b8b21b06305
14c4bd45162165
155ba8ac27b00d
hex2d793bcbca0d

49998717504013 has 2 divisors, whose sum is σ = 49998717504014. Its totient is φ = 49998717504012.

The previous prime is 49998717503983. The next prime is 49998717504083. The reversal of 49998717504013 is 31040571789994.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 49738630763844 + 260086740169 = 7052562^2 + 509987^2 .

It is a cyclic number.

It is not a de Polignac number, because 49998717504013 - 237 = 49861278550541 is a prime.

It is a super-2 number, since 2×499987175040132 (a number of 28 digits) contains 22 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 49998717504013.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (49998717504083) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 24999358752006 + 24999358752007.

It is an arithmetic number, because the mean of its divisors is an integer number (24999358752007).

Almost surely, 249998717504013 is an apocalyptic number.

It is an amenable number.

49998717504013 is a deficient number, since it is larger than the sum of its proper divisors (1).

49998717504013 is an equidigital number, since it uses as much as digits as its factorization.

49998717504013 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 68584320, while the sum is 67.

The spelling of 49998717504013 in words is "forty-nine trillion, nine hundred ninety-eight billion, seven hundred seventeen million, five hundred four thousand, thirteen".