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1202103202135 = 51147214629617
BaseRepresentation
bin10001011111100010111…
…011110011000101010111
311020220211122020211220211
4101133202323303011113
5124143401404432020
62320123311310251
7152564144424463
oct21374273630527
94226748224824
101202103202135
11423899a11a80
12174b85b1b387
1389486257204
14422799a94a3
152140968685a
hex117e2ef3157

1202103202135 has 16 divisors (see below), whose sum is σ = 1573996046112. Its totient is φ = 874071500800.

The previous prime is 1202103202133. The next prime is 1202103202151. The reversal of 1202103202135 is 5312023012021.

It is not a de Polignac number, because 1202103202135 - 21 = 1202103202133 is a prime.

It is a super-4 number, since 4×12021032021354 (a number of 49 digits) contains 4444 as substring.

It is a Duffinian number.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (1202103202133) by changing a digit.

It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 2055154 + ... + 2574463.

It is an arithmetic number, because the mean of its divisors is an integer number (98374752882).

Almost surely, 21202103202135 is an apocalyptic number.

1202103202135 is a deficient number, since it is larger than the sum of its proper divisors (371892843977).

1202103202135 is a wasteful number, since it uses less digits than its factorization.

1202103202135 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 4634354.

The product of its (nonzero) digits is 720, while the sum is 22.

Adding to 1202103202135 its reverse (5312023012021), we get a palindrome (6514126214156).

The spelling of 1202103202135 in words is "one trillion, two hundred two billion, one hundred three million, two hundred two thousand, one hundred thirty-five".

Divisors: 1 5 11 55 4721 23605 51931 259655 4629617 23148085 50925787 254628935 21856421857 109282109285 240420640427 1202103202135