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12535041550433 is a prime number
BaseRepresentation
bin1011011001101000101010…
…1100101110100001100001
31122101100010121011000110012
42312122022230232201201
53120333231114103213
642354304003214305
72432425005111443
oct266321254564141
948340117130405
1012535041550433
113aa3099192836
1214a5460077395
136cc08454b78c
143149b0a46493
1516b0ea0904a8
hexb668ab2e861

12535041550433 has 2 divisors, whose sum is σ = 12535041550434. Its totient is φ = 12535041550432.

The previous prime is 12535041550399. The next prime is 12535041550439. The reversal of 12535041550433 is 33405514053521.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 7146361799824 + 5388679750609 = 2673268^2 + 2321353^2 .

It is a cyclic number.

It is not a de Polignac number, because 12535041550433 - 232 = 12530746583137 is a prime.

It is not a weakly prime, because it can be changed into another prime (12535041550439) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6267520775216 + 6267520775217.

It is an arithmetic number, because the mean of its divisors is an integer number (6267520775217).

Almost surely, 212535041550433 is an apocalyptic number.

It is an amenable number.

12535041550433 is a deficient number, since it is larger than the sum of its proper divisors (1).

12535041550433 is an equidigital number, since it uses as much as digits as its factorization.

12535041550433 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 540000, while the sum is 41.

The spelling of 12535041550433 in words is "twelve trillion, five hundred thirty-five billion, forty-one million, five hundred fifty thousand, four hundred thirty-three".