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1451113255603 = 11131919386873
BaseRepresentation
bin10101000111011101000…
…101110001001010110011
312010201120120121012200201
4111013131011301022303
5142233334443134403
63030344321100031
7206560640104654
oct25073505611263
95121516535621
101451113255603
1150a4606697a0
121b529aa87617
13a6abb1b8c0b
145033cb2952b
1527b3060821d
hex151dd1712b3

1451113255603 has 4 divisors (see below), whose sum is σ = 1583032642488. Its totient is φ = 1319193868720.

The previous prime is 1451113255597. The next prime is 1451113255619. The reversal of 1451113255603 is 3065523111541.

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

It is a cyclic number.

It is not a de Polignac number, because 1451113255603 - 225 = 1451079701171 is a prime.

It is a super-2 number, since 2×14511132556032 (a number of 25 digits) contains 22 as substring.

It is not an unprimeable number, because it can be changed into a prime (1451113255643) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 65959693426 + ... + 65959693447.

It is an arithmetic number, because the mean of its divisors is an integer number (395758160622).

Almost surely, 21451113255603 is an apocalyptic number.

1451113255603 is a deficient number, since it is larger than the sum of its proper divisors (131919386885).

1451113255603 is a wasteful number, since it uses less digits than its factorization.

1451113255603 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 131919386884.

The product of its (nonzero) digits is 54000, while the sum is 37.

The spelling of 1451113255603 in words is "one trillion, four hundred fifty-one billion, one hundred thirteen million, two hundred fifty-five thousand, six hundred three".

Divisors: 1 11 131919386873 1451113255603