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13135412132554 = 26567706066277
BaseRepresentation
bin1011111100100101001110…
…0100101111101011001010
31201111201210110022121001011
42333021103210233223022
53210202311001220204
643534154112314134
72524000520626015
oct277112344575312
951451713277034
1013135412132554
114204773789148
121581892b4294a
13743882bba53a
14335a861d437c
1517ba37710304
hexbf25392faca

13135412132554 has 4 divisors (see below), whose sum is σ = 19703118198834. Its totient is φ = 6567706066276.

The previous prime is 13135412132537. The next prime is 13135412132581. The reversal of 13135412132554 is 45523121453131.

It is a semiprime because it is the product of two primes.

It can be written as a sum of positive squares in only one way, i.e., 7720590174025 + 5414821958529 = 2778595^2 + 2326977^2 .

It is a super-3 number, since 3×131354121325543 (a number of 40 digits) contains 333 as substring.

It is an unprimeable number.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 3283853033137 + ... + 3283853033140.

Almost surely, 213135412132554 is an apocalyptic number.

13135412132554 is a deficient number, since it is larger than the sum of its proper divisors (6567706066280).

13135412132554 is an equidigital number, since it uses as much as digits as its factorization.

13135412132554 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 6567706066279.

The product of its digits is 216000, while the sum is 40.

Adding to 13135412132554 its reverse (45523121453131), we get a palindrome (58658533585685).

The spelling of 13135412132554 in words is "thirteen trillion, one hundred thirty-five billion, four hundred twelve million, one hundred thirty-two thousand, five hundred fifty-four".

Divisors: 1 2 6567706066277 13135412132554