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11431054440050 = 2527437394111057
BaseRepresentation
bin1010011001010111111111…
…1100000111011001110010
31111110210112211122001111112
42212111333330013121302
52444241244414040200
640151204240113322
72256603120363410
oct246257774073162
944423484561445
1011431054440050
1137079792517a3
1213474b9005842
1364bc367159bb
142b73a1c222b0
1514c534563535
hexa657ff07672

11431054440050 has 192 divisors, whose sum is σ = 25233925711104. Its totient is φ = 3771281203200.

The previous prime is 11431054440019. The next prime is 11431054440121. The reversal of 11431054440050 is 5004445013411.

It is a super-2 number, since 2×114310544400502 (a number of 27 digits) contains 22 as substring.

It is a Curzon number.

It is a junction number, because it is equal to n+sod(n) for n = 11431054439989 and 11431054440016.

It is an unprimeable number.

It is a polite number, since it can be written in 95 ways as a sum of consecutive naturals, for example, 1033824122 + ... + 1033835178.

It is an arithmetic number, because the mean of its divisors is an integer number (131426696412).

Almost surely, 211431054440050 is an apocalyptic number.

11431054440050 is a gapful number since it is divisible by the number (10) formed by its first and last digit.

11431054440050 is an abundant number, since it is smaller than the sum of its proper divisors (13802871271054).

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

11431054440050 is a wasteful number, since it uses less digits than its factorization.

11431054440050 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 12133 (or 12128 counting only the distinct ones).

The product of its (nonzero) digits is 19200, while the sum is 32.

Adding to 11431054440050 its reverse (5004445013411), we get a palindrome (16435499453461).

The spelling of 11431054440050 in words is "eleven trillion, four hundred thirty-one billion, fifty-four million, four hundred forty thousand, fifty".