Base | Representation |
---|---|
bin | 100010101010100110011101… |
… | …101101110001111011111111 |
3 | 201222211010202012120012211222 |
4 | 202222212131231301323333 |
5 | 124440410313113130142 |
6 | 1300131355033103555 |
7 | 44053644121534412 |
oct | 4252463555617377 |
9 | 658733665505758 |
10 | 152461100130047 |
11 | 44640431409495 |
12 | 15123b95b57bbb |
13 | 670c035093707 |
14 | 2991229a95b79 |
15 | 1295ce2e17bd2 |
hex | 8aa99db71eff |
152461100130047 has 2 divisors, whose sum is σ = 152461100130048. Its totient is φ = 152461100130046.
The previous prime is 152461100130019. The next prime is 152461100130083. The reversal of 152461100130047 is 740031001164251.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 152461100130047 - 28 = 152461100129791 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 152461100129989 and 152461100130016.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (152461100130247) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 76230550065023 + 76230550065024.
It is an arithmetic number, because the mean of its divisors is an integer number (76230550065024).
Almost surely, 2152461100130047 is an apocalyptic number.
152461100130047 is a deficient number, since it is larger than the sum of its proper divisors (1).
152461100130047 is an equidigital number, since it uses as much as digits as its factorization.
152461100130047 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 20160, while the sum is 35.
Adding to 152461100130047 its reverse (740031001164251), we get a palindrome (892492101294298).
The spelling of 152461100130047 in words is "one hundred fifty-two trillion, four hundred sixty-one billion, one hundred million, one hundred thirty thousand, forty-seven".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.075 sec. • engine limits •