Base | Representation |
---|---|
bin | 10011000100000100000… |
… | …001001001010100010011 |
3 | 11122020102102010221001111 |
4 | 103010010001021110103 |
5 | 132430421300400301 |
6 | 2441453125314151 |
7 | 163434552651406 |
oct | 23040401112423 |
9 | 4566372127044 |
10 | 1310032434451 |
11 | 46564417a734 |
12 | 191a87265957 |
13 | 966c672c9a0 |
14 | 47597b51d3d |
15 | 24124a8d551 |
hex | 13104049513 |
1310032434451 has 12 divisors (see below), whose sum is σ = 1430395061088. Its totient is φ = 1192695430464.
The previous prime is 1310032434449. The next prime is 1310032434457. The reversal of 1310032434451 is 1544342300131.
It is not a de Polignac number, because 1310032434451 - 21 = 1310032434449 is a prime.
It is a super-2 number, since 2×13100324344512 (a number of 25 digits) contains 22 as substring.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (1310032434457) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (13) of ones.
It is a polite number, since it can be written in 11 ways as a sum of consecutive naturals, for example, 9385755 + ... + 9524308.
It is an arithmetic number, because the mean of its divisors is an integer number (119199588424).
Almost surely, 21310032434451 is an apocalyptic number.
1310032434451 is a deficient number, since it is larger than the sum of its proper divisors (120362626637).
1310032434451 is an equidigital number, since it uses as much as digits as its factorization.
1310032434451 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 18910222 (or 18910149 counting only the distinct ones).
The product of its (nonzero) digits is 17280, while the sum is 31.
Adding to 1310032434451 its reverse (1544342300131), we get a palindrome (2854374734582).
The spelling of 1310032434451 in words is "one trillion, three hundred ten billion, thirty-two million, four hundred thirty-four thousand, four hundred fifty-one".
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