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67994353153 = 103660139351
BaseRepresentation
bin111111010100110001…
…110111111000000001
320111111122101000122001
4333110301313320001
52103223023300103
651123002544001
74624651315516
oct772461677001
9214448330561
1067994353153
11269220489a1
1211217208001
136547a73c7c
1434104cb30d
151b7e4d501d
hexfd4c77e01

67994353153 has 4 divisors (see below), whose sum is σ = 68654492608. Its totient is φ = 67334213700.

The previous prime is 67994353123. The next prime is 67994353159. The reversal of 67994353153 is 35135349976.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.

It is a cyclic number.

It is not a de Polignac number, because 67994353153 - 25 = 67994353121 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 67994353094 and 67994353103.

It is not an unprimeable number, because it can be changed into a prime (67994353159) 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, 330069573 + ... + 330069778.

It is an arithmetic number, because the mean of its divisors is an integer number (17163623152).

Almost surely, 267994353153 is an apocalyptic number.

It is an amenable number.

67994353153 is a deficient number, since it is larger than the sum of its proper divisors (660139455).

67994353153 is a wasteful number, since it uses less digits than its factorization.

67994353153 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 660139454.

The product of its digits is 9185400, while the sum is 55.

The spelling of 67994353153 in words is "sixty-seven billion, nine hundred ninety-four million, three hundred fifty-three thousand, one hundred fifty-three".

Divisors: 1 103 660139351 67994353153