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49999861239413 is a prime number
BaseRepresentation
bin10110101111001011111111…
…11101111100111001110101
320120000221101022001111101212
423113211333331330321311
523023144203434130123
6254201341423131205
713350241113112241
oct1327457775747165
9216027338044355
1049999861239413
1114a279051a3218
125736384443b05
1321b8c74a5ab17
14c4c013007421
155ba9278a0878
hex2d797ff7ce75

49999861239413 has 2 divisors, whose sum is σ = 49999861239414. Its totient is φ = 49999861239412.

The previous prime is 49999861239359. The next prime is 49999861239431. The reversal of 49999861239413 is 31493216899994.

Together with next prime (49999861239431) it forms an Ormiston pair, because they use the same digits, order apart.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 49999861239364 + 49 = 7071058^2 + 7^2 .

It is a cyclic number.

It is not a de Polignac number, because 49999861239413 - 232 = 49995566272117 is a prime.

It is a super-2 number, since 2×499998612394132 (a number of 28 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (49999861209413) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 24999930619706 + 24999930619707.

It is an arithmetic number, because the mean of its divisors is an integer number (24999930619707).

Almost surely, 249999861239413 is an apocalyptic number.

It is an amenable number.

49999861239413 is a deficient number, since it is larger than the sum of its proper divisors (1).

49999861239413 is an equidigital number, since it uses as much as digits as its factorization.

49999861239413 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 816293376, while the sum is 77.

The spelling of 49999861239413 in words is "forty-nine trillion, nine hundred ninety-nine billion, eight hundred sixty-one million, two hundred thirty-nine thousand, four hundred thirteen".