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131102224531 is a prime number
BaseRepresentation
bin111101000011001001…
…1010011000010010011
3110112101201222001102121
41322012103103002103
54121444132141111
6140121051355111
712320556343645
oct1720623230223
9415351861377
10131102224531
11506668737a3
12214a9a52a97
13c494353bb5
1464b9a26895
1536249d8871
hex1e864d3093

131102224531 has 2 divisors, whose sum is σ = 131102224532. Its totient is φ = 131102224530.

The previous prime is 131102224507. The next prime is 131102224541. The reversal of 131102224531 is 135422201131.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 131102224531 - 213 = 131102216339 is a prime.

It is a super-2 number, since 2×1311022245312 (a number of 23 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 131102224496 and 131102224505.

It is not a weakly prime, because it can be changed into another prime (131102224541) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65551112265 + 65551112266.

It is an arithmetic number, because the mean of its divisors is an integer number (65551112266).

Almost surely, 2131102224531 is an apocalyptic number.

131102224531 is a deficient number, since it is larger than the sum of its proper divisors (1).

131102224531 is an equidigital number, since it uses as much as digits as its factorization.

131102224531 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 1440, while the sum is 25.

Adding to 131102224531 its reverse (135422201131), we get a palindrome (266524425662).

The spelling of 131102224531 in words is "one hundred thirty-one billion, one hundred two million, two hundred twenty-four thousand, five hundred thirty-one".