Base | Representation |
---|---|
bin | 11100001111110110000011… |
… | …110010100101111000000001 |
3 | 121021212200000211101221001201 |
4 | 130033312003302211320001 |
5 | 112240423004332130213 |
6 | 1120120213122101201 |
7 | 35111422543264516 |
oct | 3417660362457001 |
9 | 537780024357051 |
10 | 124234140114433 |
11 | 36648448354288 |
12 | 11b254aa1aa801 |
13 | 54422c95784cb |
14 | 2296d706c330d |
15 | e56935dae6dd |
hex | 70fd83ca5e01 |
124234140114433 has 2 divisors, whose sum is σ = 124234140114434. Its totient is φ = 124234140114432.
The previous prime is 124234140114431. The next prime is 124234140114437. The reversal of 124234140114433 is 334411041432421.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 108437712435649 + 15796427678784 = 10413343^2 + 3974472^2 .
It is a cyclic number.
It is not a de Polignac number, because 124234140114433 - 21 = 124234140114431 is a prime.
It is a super-2 number, since 2×1242341401144332 (a number of 29 digits) contains 22 as substring.
Together with 124234140114431, it forms a pair of twin primes.
It is a junction number, because it is equal to n+sod(n) for n = 124234140114392 and 124234140114401.
It is not a weakly prime, because it can be changed into another prime (124234140114431) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 62117070057216 + 62117070057217.
It is an arithmetic number, because the mean of its divisors is an integer number (62117070057217).
Almost surely, 2124234140114433 is an apocalyptic number.
It is an amenable number.
124234140114433 is a deficient number, since it is larger than the sum of its proper divisors (1).
124234140114433 is an equidigital number, since it uses as much as digits as its factorization.
124234140114433 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 110592, while the sum is 37.
Adding to 124234140114433 its reverse (334411041432421), we get a palindrome (458645181546854).
The spelling of 124234140114433 in words is "one hundred twenty-four trillion, two hundred thirty-four billion, one hundred forty million, one hundred fourteen thousand, four hundred thirty-three".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.094 sec. • engine limits •