Base | Representation |
---|---|
bin | 10000101111111101100… |
… | …010101100111000101111 |
3 | 11002000221212101021110222 |
4 | 100233331202230320233 |
5 | 122324232030003211 |
6 | 2240433334521555 |
7 | 146105046401654 |
oct | 20577542547057 |
9 | 4060855337428 |
10 | 1151010000431 |
11 | 404160158345 |
12 | 1670a6b062bb |
13 | 84702b76c95 |
14 | 3d9cddc9c2b |
15 | 1ee18d501db |
hex | 10bfd8ace2f |
1151010000431 has 2 divisors, whose sum is σ = 1151010000432. Its totient is φ = 1151010000430.
The previous prime is 1151010000409. The next prime is 1151010000487. The reversal of 1151010000431 is 1340000101511.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 1151010000431 - 218 = 1151009738287 is a prime.
It is a super-2 number, since 2×11510100004312 (a number of 25 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1151010000131) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 575505000215 + 575505000216.
It is an arithmetic number, because the mean of its divisors is an integer number (575505000216).
Almost surely, 21151010000431 is an apocalyptic number.
1151010000431 is a deficient number, since it is larger than the sum of its proper divisors (1).
1151010000431 is an equidigital number, since it uses as much as digits as its factorization.
1151010000431 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 60, while the sum is 17.
Adding to 1151010000431 its reverse (1340000101511), we get a palindrome (2491010101942).
The spelling of 1151010000431 in words is "one trillion, one hundred fifty-one billion, ten million, four hundred thirty-one", and thus it is an aban number.
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