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123780504952667 is a prime number
BaseRepresentation
bin11100001001001111100101…
…000001110001111101011011
3121020021022010012211000222002
4130021033211001301331123
5112211004433231441132
61115131551242151215
735033563206111503
oct3411174501617533
9536238105730862
10123780504952667
1136493021105012
1211a715a87b250b
13540b5b4028a74
14227d0190caa03
15e49c35916262
hex7093e5071f5b

123780504952667 has 2 divisors, whose sum is σ = 123780504952668. Its totient is φ = 123780504952666.

The previous prime is 123780504952649. The next prime is 123780504952669. The reversal of 123780504952667 is 766259405087321.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 123780504952667 - 28 = 123780504952411 is a prime.

It is a super-2 number, since 2×1237805049526672 (a number of 29 digits) contains 22 as substring.

Together with 123780504952669, it forms a pair of twin primes.

It is a Chen prime.

It is not a weakly prime, because it can be changed into another prime (123780504952669) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 61890252476333 + 61890252476334.

It is an arithmetic number, because the mean of its divisors is an integer number (61890252476334).

It is a 1-persistent number, because it is pandigital, but 2⋅123780504952667 = 247561009905334 is not.

Almost surely, 2123780504952667 is an apocalyptic number.

123780504952667 is a deficient number, since it is larger than the sum of its proper divisors (1).

123780504952667 is an equidigital number, since it uses as much as digits as its factorization.

123780504952667 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 152409600, while the sum is 65.

The spelling of 123780504952667 in words is "one hundred twenty-three trillion, seven hundred eighty billion, five hundred four million, nine hundred fifty-two thousand, six hundred sixty-seven".