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311354202453114 = 23192731177214501
BaseRepresentation
bin100011011001011001100110…
…0101000101111010001111010
31111211102011121201000012110010
41012302303030220233101322
5311302211401311444424
63022110051311402350
7122403412250100522
oct10662631450572172
91454364551005403
10311354202453114
119023063732aa27
122ab066445233b6
1310496762c51922
1456c58a3a14682
1525ee085e83229
hex11b2ccca2f47a

311354202453114 has 16 divisors (see below), whose sum is σ = 655482531480480. Its totient is φ = 98322379722000.

The previous prime is 311354202453101. The next prime is 311354202453181. The reversal of 311354202453114 is 411354202453113.

It is a super-3 number, since 3×3113542024531143 (a number of 44 digits) contains 333 as substring.

It is a Curzon number.

It is an unprimeable number.

It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 1365588607137 + ... + 1365588607364.

It is an arithmetic number, because the mean of its divisors is an integer number (40967658217530).

Almost surely, 2311354202453114 is an apocalyptic number.

311354202453114 is an abundant number, since it is smaller than the sum of its proper divisors (344128329027366).

It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.

311354202453114 is a wasteful number, since it uses less digits than its factorization.

311354202453114 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 2731177214525.

The product of its (nonzero) digits is 172800, while the sum is 39.

Subtracting 311354202453114 from its reverse (411354202453113), we obtain a palindrome (99999999999999).

The spelling of 311354202453114 in words is "three hundred eleven trillion, three hundred fifty-four billion, two hundred two million, four hundred fifty-three thousand, one hundred fourteen".

Divisors: 1 2 3 6 19 38 57 114 2731177214501 5462354429002 8193531643503 16387063287006 51892367075519 103784734151038 155677101226557 311354202453114