Base | Representation |
---|---|
bin | 10011101101101011010… |
… | …000101110000011110001 |
3 | 11210111202100211222221012 |
4 | 103231223100232003301 |
5 | 134143423321241223 |
6 | 2514202551440305 |
7 | 166606040033624 |
oct | 23555320560361 |
9 | 4714670758835 |
10 | 1354714243313 |
11 | 482592966aa2 |
12 | 19a677086095 |
13 | 9a997744042 |
14 | 497d6033dbb |
15 | 2538c5c7c78 |
hex | 13b6b42e0f1 |
1354714243313 has 2 divisors, whose sum is σ = 1354714243314. Its totient is φ = 1354714243312.
The previous prime is 1354714243303. The next prime is 1354714243321. The reversal of 1354714243313 is 3133424174531.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 1124422710544 + 230291532769 = 1060388^2 + 479887^2 .
It is a cyclic number.
It is not a de Polignac number, because 1354714243313 - 230 = 1353640501489 is a prime.
It is a super-2 number, since 2×13547142433132 (a number of 25 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (1354714243303) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 677357121656 + 677357121657.
It is an arithmetic number, because the mean of its divisors is an integer number (677357121657).
Almost surely, 21354714243313 is an apocalyptic number.
It is an amenable number.
1354714243313 is a deficient number, since it is larger than the sum of its proper divisors (1).
1354714243313 is an equidigital number, since it uses as much as digits as its factorization.
1354714243313 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 362880, while the sum is 41.
The spelling of 1354714243313 in words is "one trillion, three hundred fifty-four billion, seven hundred fourteen million, two hundred forty-three thousand, three hundred thirteen".
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