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1434643390751 is a prime number
BaseRepresentation
bin10100111000000111011…
…010001110000100011111
312002011001202122101111012
4110320013122032010133
5142001122202001001
63015022134445435
7205435541421164
oct24700732160437
95064052571435
101434643390751
1150347987a53a
121b206320387b
13a5394c9c25c
144d61960806b
15274b974d2bb
hex14e0768e11f

1434643390751 has 2 divisors, whose sum is σ = 1434643390752. Its totient is φ = 1434643390750.

The previous prime is 1434643390723. The next prime is 1434643390871. The reversal of 1434643390751 is 1570933464341.

It is a weak prime.

It is an emirp because it is prime and its reverse (1570933464341) is a distict prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-1434643390751 is a prime.

It is a super-3 number, since 3×14346433907513 (a number of 37 digits) contains 333 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 1434643390751.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1434643399751) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 717321695375 + 717321695376.

It is an arithmetic number, because the mean of its divisors is an integer number (717321695376).

Almost surely, 21434643390751 is an apocalyptic number.

1434643390751 is a deficient number, since it is larger than the sum of its proper divisors (1).

1434643390751 is an equidigital number, since it uses as much as digits as its factorization.

1434643390751 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 3265920, while the sum is 50.

The spelling of 1434643390751 in words is "one trillion, four hundred thirty-four billion, six hundred forty-three million, three hundred ninety thousand, seven hundred fifty-one".