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13442040043303 is a prime number
BaseRepresentation
bin1100001110011011100000…
…0001011001011100100111
31202121001020212211101002011
43003212320001121130213
53230213244222341203
644331104332455051
72555104166103334
oct303467001313447
952531225741064
1013442040043303
114312812197933
1216111a801a487
1376676ac03811
143468536b598b
151849d1bd976d
hexc39b8059727

13442040043303 has 2 divisors, whose sum is σ = 13442040043304. Its totient is φ = 13442040043302.

The previous prime is 13442040043273. The next prime is 13442040043361. The reversal of 13442040043303 is 30334004024431.

It is a weak prime.

It is an emirp because it is prime and its reverse (30334004024431) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 13442040043303 - 29 = 13442040042791 is a prime.

It is a super-2 number, since 2×134420400433032 (a number of 27 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (13442040043903) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6721020021651 + 6721020021652.

It is an arithmetic number, because the mean of its divisors is an integer number (6721020021652).

Almost surely, 213442040043303 is an apocalyptic number.

13442040043303 is a deficient number, since it is larger than the sum of its proper divisors (1).

13442040043303 is an equidigital number, since it uses as much as digits as its factorization.

13442040043303 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 41472, while the sum is 31.

Adding to 13442040043303 its reverse (30334004024431), we get a palindrome (43776044067734).

The spelling of 13442040043303 in words is "thirteen trillion, four hundred forty-two billion, forty million, forty-three thousand, three hundred three".