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13354454424169 is a prime number
BaseRepresentation
bin1100001001010101001110…
…0000110001101001101001
31202021200011220000220212101
43002111103200301221221
53222244410313033134
644222541321013401
72545553554450135
oct302252340615151
952250156026771
1013354454424169
114289657410931
1215b8223924261
1375b420363a91
143425052dc3c5
151825a77c4e14
hexc2553831a69

13354454424169 has 2 divisors, whose sum is σ = 13354454424170. Its totient is φ = 13354454424168.

The previous prime is 13354454424089. The next prime is 13354454424191. The reversal of 13354454424169 is 96142445445331.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 11854331632144 + 1500122792025 = 3443012^2 + 1224795^2 .

It is a cyclic number.

It is not a de Polignac number, because 13354454424169 - 231 = 13352306940521 is a prime.

It is a super-2 number, since 2×133544544241692 (a number of 27 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (13354454424569) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6677227212084 + 6677227212085.

It is an arithmetic number, because the mean of its divisors is an integer number (6677227212085).

Almost surely, 213354454424169 is an apocalyptic number.

It is an amenable number.

13354454424169 is a deficient number, since it is larger than the sum of its proper divisors (1).

13354454424169 is an equidigital number, since it uses as much as digits as its factorization.

13354454424169 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 24883200, while the sum is 55.

The spelling of 13354454424169 in words is "thirteen trillion, three hundred fifty-four billion, four hundred fifty-four million, four hundred twenty-four thousand, one hundred sixty-nine".