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87367680433 = 117942516403
BaseRepresentation
bin101000101011110000…
…1010000000110110001
322100111210201000112001
41101113201100012301
52412412111233213
6104045224334001
76212024013325
oct1212741200661
9270453630461
1087367680433
1134063861480
1214b23314301
138314674637
14432b485985
1524152151dd
hex14578501b1

87367680433 has 4 divisors (see below), whose sum is σ = 95310196848. Its totient is φ = 79425164020.

The previous prime is 87367680431. The next prime is 87367680451. The reversal of 87367680433 is 33408676378.

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 87367680433 - 21 = 87367680431 is a prime.

It is a super-2 number, since 2×873676804332 (a number of 23 digits) contains 22 as substring.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (87367680431) 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, 3971258191 + ... + 3971258212.

It is an arithmetic number, because the mean of its divisors is an integer number (23827549212).

Almost surely, 287367680433 is an apocalyptic number.

It is an amenable number.

87367680433 is a deficient number, since it is larger than the sum of its proper divisors (7942516415).

87367680433 is a wasteful number, since it uses less digits than its factorization.

87367680433 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 7942516414.

The product of its (nonzero) digits is 12192768, while the sum is 55.

The spelling of 87367680433 in words is "eighty-seven billion, three hundred sixty-seven million, six hundred eighty thousand, four hundred thirty-three".

Divisors: 1 11 7942516403 87367680433