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13101721506745 = 52620344301349
BaseRepresentation
bin1011111010100111101101…
…1101001101001110111001
31201101111211111110110211021
42332221323131031032321
53204124311201203440
643510503041302441
72521365605215114
oct276517335151671
951344744413737
1013101721506745
1141a1455258343
121577250034a21
137406440345b1
143341ab96957b
1517ac14ac284a
hexbea7b74d3b9

13101721506745 has 4 divisors (see below), whose sum is σ = 15722065808100. Its totient is φ = 10481377205392.

The previous prime is 13101721506719. The next prime is 13101721506803. The reversal of 13101721506745 is 54760512710131.

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

It can be written as a sum of positive squares in 2 ways, for example, as 4941146591161 + 8160574915584 = 2222869^2 + 2856672^2 .

It is a cyclic number.

It is not a de Polignac number, because 13101721506745 - 213 = 13101721498553 is a prime.

It is a super-3 number, since 3×131017215067453 (a number of 40 digits) contains 333 as substring.

It is an unprimeable number.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1310172150670 + ... + 1310172150679.

It is an arithmetic number, because the mean of its divisors is an integer number (3930516452025).

Almost surely, 213101721506745 is an apocalyptic number.

It is an amenable number.

13101721506745 is a deficient number, since it is larger than the sum of its proper divisors (2620344301355).

13101721506745 is an equidigital number, since it uses as much as digits as its factorization.

13101721506745 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 2620344301354.

The product of its (nonzero) digits is 176400, while the sum is 43.

The spelling of 13101721506745 in words is "thirteen trillion, one hundred one billion, seven hundred twenty-one million, five hundred six thousand, seven hundred forty-five".

Divisors: 1 5 2620344301349 13101721506745