Base | Representation |
---|---|
bin | 11101110100010110101011… |
… | …010110010011101001100101 |
3 | 122012022222010112001102022211 |
4 | 131310112223112103221211 |
5 | 114142103443431402412 |
6 | 1142525221135123421 |
7 | 36423426662652136 |
oct | 3564265326235145 |
9 | 565288115042284 |
10 | 131141111200357 |
11 | 388706a6481565 |
12 | 12860038973571 |
13 | 58237276b56cb |
14 | 2455399649c8d |
15 | 102643437aca7 |
hex | 7745ab593a65 |
131141111200357 has 2 divisors, whose sum is σ = 131141111200358. Its totient is φ = 131141111200356.
The previous prime is 131141111200283. The next prime is 131141111200397. The reversal of 131141111200357 is 753002111141131.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 119563159036036 + 11577952164321 = 10934494^2 + 3402639^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-131141111200357 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (131141111200397) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65570555600178 + 65570555600179.
It is an arithmetic number, because the mean of its divisors is an integer number (65570555600179).
Almost surely, 2131141111200357 is an apocalyptic number.
It is an amenable number.
131141111200357 is a deficient number, since it is larger than the sum of its proper divisors (1).
131141111200357 is an equidigital number, since it uses as much as digits as its factorization.
131141111200357 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 2520, while the sum is 31.
Adding to 131141111200357 its reverse (753002111141131), we get a palindrome (884143222341488).
The spelling of 131141111200357 in words is "one hundred thirty-one trillion, one hundred forty-one billion, one hundred eleven million, two hundred thousand, three hundred fifty-seven".
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