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13421021405573 is a prime number
BaseRepresentation
bin1100001101001101001100…
…1101101010000110000101
31202112001000001112221120002
43003103103031222012011
53224342222434434243
644313310534251045
72553431264452205
oct303232315520605
952461001487502
1013421021405573
114304906775251
121609100ab6a85
1376479b494682
1434581bdd0405
151841a1809eb8
hexc34d336a185

13421021405573 has 2 divisors, whose sum is σ = 13421021405574. Its totient is φ = 13421021405572.

The previous prime is 13421021405503. The next prime is 13421021405593. The reversal of 13421021405573 is 37550412012431.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 12849306975649 + 571714429924 = 3584593^2 + 756118^2 .

It is a cyclic number.

It is not a de Polignac number, because 13421021405573 - 214 = 13421021389189 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (13421021405503) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6710510702786 + 6710510702787.

It is an arithmetic number, because the mean of its divisors is an integer number (6710510702787).

Almost surely, 213421021405573 is an apocalyptic number.

It is an amenable number.

13421021405573 is a deficient number, since it is larger than the sum of its proper divisors (1).

13421021405573 is an equidigital number, since it uses as much as digits as its factorization.

13421021405573 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 100800, while the sum is 38.

The spelling of 13421021405573 in words is "thirteen trillion, four hundred twenty-one billion, twenty-one million, four hundred five thousand, five hundred seventy-three".