Base | Representation |
---|---|
bin | 10111011100110000101111… |
… | …101101010101000101011111 |
3 | 111112011021120101202111211012 |
4 | 113130300233231111011133 |
5 | 102004201411102423201 |
6 | 1003201552105045435 |
7 | 30503003035405523 |
oct | 2734605755250537 |
9 | 445137511674735 |
10 | 103131555123551 |
11 | 2a951935262359 |
12 | b69770518687b |
13 | 4571357aa0106 |
14 | 1b67845945d83 |
15 | bdca4d65e5bb |
hex | 5dcc2fb5515f |
103131555123551 has 2 divisors, whose sum is σ = 103131555123552. Its totient is φ = 103131555123550.
The previous prime is 103131555123461. The next prime is 103131555123577. The reversal of 103131555123551 is 155321555131301.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 103131555123551 - 230 = 103130481381727 is a prime.
It is a super-2 number, since 2×1031315551235512 (a number of 29 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 103131555123499 and 103131555123508.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (103131555122551) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 51565777561775 + 51565777561776.
It is an arithmetic number, because the mean of its divisors is an integer number (51565777561776).
Almost surely, 2103131555123551 is an apocalyptic number.
103131555123551 is a deficient number, since it is larger than the sum of its proper divisors (1).
103131555123551 is an equidigital number, since it uses as much as digits as its factorization.
103131555123551 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 168750, while the sum is 41.
The spelling of 103131555123551 in words is "one hundred three trillion, one hundred thirty-one billion, five hundred fifty-five million, one hundred twenty-three thousand, five hundred fifty-one".
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