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31139974477 is a prime number
BaseRepresentation
bin11101000000000101…
…100100110101001101
32222101012100121101001
4131000011210311031
51002233313140402
622145553420301
72151451262413
oct350005446515
988335317331
1031139974477
111222a7817a7
12605086a691
132c23600b01
1417159d12b3
15c23c59187
hex740164d4d

31139974477 has 2 divisors, whose sum is σ = 31139974478. Its totient is φ = 31139974476.

The previous prime is 31139974457. The next prime is 31139974483. The reversal of 31139974477 is 77447993113.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 30343549636 + 796424841 = 174194^2 + 28221^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-31139974477 is a prime.

It is a super-2 number, since 2×311399744772 (a number of 22 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (31139974457) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 15569987238 + 15569987239.

It is an arithmetic number, because the mean of its divisors is an integer number (15569987239).

Almost surely, 231139974477 is an apocalyptic number.

It is an amenable number.

31139974477 is a deficient number, since it is larger than the sum of its proper divisors (1).

31139974477 is an equidigital number, since it uses as much as digits as its factorization.

31139974477 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 4000752, while the sum is 55.

The spelling of 31139974477 in words is "thirty-one billion, one hundred thirty-nine million, nine hundred seventy-four thousand, four hundred seventy-seven".