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43133343102433 is a prime number
BaseRepresentation
bin10011100111010110000111…
…00101111111010111100001
312122201111200211202201210001
421303223003211333113201
521123144041313234213
6231423103445140001
712041165311006363
oct1163530345772741
9178644624681701
1043133343102433
111281a825774251
124a076453ba601
131b0b5c27931bc
14a91944876d33
154ebee5a3d1dd
hex273ac397f5e1

43133343102433 has 2 divisors, whose sum is σ = 43133343102434. Its totient is φ = 43133343102432.

The previous prime is 43133343102427. The next prime is 43133343102461. The reversal of 43133343102433 is 33420134333134.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 35787817573264 + 7345525529169 = 5982292^2 + 2710263^2 .

It is a cyclic number.

It is not a de Polignac number, because 43133343102433 - 29 = 43133343101921 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 43133343102392 and 43133343102401.

It is not a weakly prime, because it can be changed into another prime (43133343102233) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 21566671551216 + 21566671551217.

It is an arithmetic number, because the mean of its divisors is an integer number (21566671551217).

Almost surely, 243133343102433 is an apocalyptic number.

It is an amenable number.

43133343102433 is a deficient number, since it is larger than the sum of its proper divisors (1).

43133343102433 is an equidigital number, since it uses as much as digits as its factorization.

43133343102433 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 279936, while the sum is 37.

Adding to 43133343102433 its reverse (33420134333134), we get a palindrome (76553477435567).

The spelling of 43133343102433 in words is "forty-three trillion, one hundred thirty-three billion, three hundred forty-three million, one hundred two thousand, four hundred thirty-three".