Base | Representation |
---|---|
bin | 10110100110011100001… |
… | …111010000101101000011 |
3 | 12111110211112000012121112 |
4 | 112212130033100231003 |
5 | 200421224214041011 |
6 | 3145253014113535 |
7 | 220131242631116 |
oct | 26463417205503 |
9 | 5443745005545 |
10 | 1553104440131 |
11 | 549738a70242 |
12 | 21100364b8ab |
13 | b35c33a734a |
14 | 552563cc57d |
15 | 2a5ee56358b |
hex | 1699c3d0b43 |
1553104440131 has 2 divisors, whose sum is σ = 1553104440132. Its totient is φ = 1553104440130.
The previous prime is 1553104440091. The next prime is 1553104440169. The reversal of 1553104440131 is 1310444013551.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1553104440131 is a prime.
It is a super-2 number, since 2×15531044401312 (a number of 25 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (1553104443131) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 776552220065 + 776552220066.
It is an arithmetic number, because the mean of its divisors is an integer number (776552220066).
Almost surely, 21553104440131 is an apocalyptic number.
1553104440131 is a deficient number, since it is larger than the sum of its proper divisors (1).
1553104440131 is an equidigital number, since it uses as much as digits as its factorization.
1553104440131 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 14400, while the sum is 32.
Adding to 1553104440131 its reverse (1310444013551), we get a palindrome (2863548453682).
The spelling of 1553104440131 in words is "one trillion, five hundred fifty-three billion, one hundred four million, four hundred forty thousand, one hundred thirty-one".
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