Base | Representation |
---|---|
bin | 10011101001101010010… |
… | …011001000101010000011 |
3 | 11210002121220201212001010 |
4 | 103221222103020222003 |
5 | 134111111200211120 |
6 | 2512211105412003 |
7 | 166364151142236 |
oct | 23515223105203 |
9 | 4702556655033 |
10 | 1350403132035 |
11 | 4807803aa311 |
12 | 19987333a003 |
13 | 9a45b534219 |
14 | 4950765741d |
15 | 251d8d9b7e0 |
hex | 13a6a4c8a83 |
1350403132035 has 64 divisors (see below), whose sum is σ = 2309628764160. Its totient is φ = 671372943360.
The previous prime is 1350403131977. The next prime is 1350403132063. The reversal of 1350403132035 is 5302313040531.
It is not a de Polignac number, because 1350403132035 - 26 = 1350403131971 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 1350403131993 and 1350403132011.
It is an unprimeable number.
It is a polite number, since it can be written in 63 ways as a sum of consecutive naturals, for example, 42370936 + ... + 42402794.
It is an arithmetic number, because the mean of its divisors is an integer number (36087949440).
Almost surely, 21350403132035 is an apocalyptic number.
1350403132035 is a gapful number since it is divisible by the number (15) formed by its first and last digit.
1350403132035 is a deficient number, since it is larger than the sum of its proper divisors (959225632125).
1350403132035 is a wasteful number, since it uses less digits than its factorization.
1350403132035 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 33468.
The product of its (nonzero) digits is 16200, while the sum is 30.
Adding to 1350403132035 its reverse (5302313040531), we get a palindrome (6652716172566).
The spelling of 1350403132035 in words is "one trillion, three hundred fifty billion, four hundred three million, one hundred thirty-two thousand, thirty-five".
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