Base | Representation |
---|---|
bin | 1000110011101000110… |
… | …1101100010110001001 |
3 | 112110112101121221120221 |
4 | 2030322031230112021 |
5 | 4434330413222101 |
6 | 153301223150041 |
7 | 13634240141446 |
oct | 2147215542611 |
9 | 473471557527 |
10 | 151300523401 |
11 | 59191223685 |
12 | 253a62a5321 |
13 | 11362b3b731 |
14 | 74743aa9cd |
15 | 3e07d74aa1 |
hex | 233a36c589 |
151300523401 has 2 divisors, whose sum is σ = 151300523402. Its totient is φ = 151300523400.
The previous prime is 151300523369. The next prime is 151300523419. The reversal of 151300523401 is 104325003151.
It is a happy number.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 150776113401 + 524410000 = 388299^2 + 22900^2 .
It is an emirp because it is prime and its reverse (104325003151) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 151300523401 - 25 = 151300523369 is a prime.
It is not a weakly prime, because it can be changed into another prime (151300523461) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 75650261700 + 75650261701.
It is an arithmetic number, because the mean of its divisors is an integer number (75650261701).
Almost surely, 2151300523401 is an apocalyptic number.
It is an amenable number.
151300523401 is a deficient number, since it is larger than the sum of its proper divisors (1).
151300523401 is an equidigital number, since it uses as much as digits as its factorization.
151300523401 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1800, while the sum is 25.
Adding to 151300523401 its reverse (104325003151), we get a palindrome (255625526552).
The spelling of 151300523401 in words is "one hundred fifty-one billion, three hundred million, five hundred twenty-three thousand, four hundred one".
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