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40679399141 is a prime number
BaseRepresentation
bin100101111000101011…
…100101111011100101
310220000000110120220122
4211320223211323211
51131302411233031
630404324411325
72640034121213
oct457053457345
9126000416818
1040679399141
1116285507737
127a73546b45
133ab3a5aba4
141d7c901cb3
1510d148a47b
hex978ae5ee5

40679399141 has 2 divisors, whose sum is σ = 40679399142. Its totient is φ = 40679399140.

The previous prime is 40679399113. The next prime is 40679399143. The reversal of 40679399141 is 14199397604.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 25943867041 + 14735532100 = 161071^2 + 121390^2 .

It is a cyclic number.

It is not a de Polignac number, because 40679399141 - 218 = 40679136997 is a prime.

Together with 40679399143, it forms a pair of twin primes.

It is a Chen prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (40679399143) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 20339699570 + 20339699571.

It is an arithmetic number, because the mean of its divisors is an integer number (20339699571).

Almost surely, 240679399141 is an apocalyptic number.

It is an amenable number.

40679399141 is a deficient number, since it is larger than the sum of its proper divisors (1).

40679399141 is an equidigital number, since it uses as much as digits as its factorization.

40679399141 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 1469664, while the sum is 53.

The spelling of 40679399141 in words is "forty billion, six hundred seventy-nine million, three hundred ninety-nine thousand, one hundred forty-one".