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131344422550151 is a prime number
BaseRepresentation
bin11101110111010100000001…
…101001100011101010000111
3122020001101220122020002210012
4131313110001221203222013
5114203421334133101101
61143202443355020435
736444215142632156
oct3567240151435207
9566041818202705
10131344422550151
1138939947429888
121289351943a71b
135839958a5bc55
14246116541d59d
15102b883395bbb
hex777501a63a87

131344422550151 has 2 divisors, whose sum is σ = 131344422550152. Its totient is φ = 131344422550150.

The previous prime is 131344422550147. The next prime is 131344422550157. The reversal of 131344422550151 is 151055224443131.

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 131344422550151 - 22 = 131344422550147 is a prime.

It is a super-2 number, since 2×1313444225501512 (a number of 29 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 131344422550099 and 131344422550108.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (131344422550157) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65672211275075 + 65672211275076.

It is an arithmetic number, because the mean of its divisors is an integer number (65672211275076).

Almost surely, 2131344422550151 is an apocalyptic number.

131344422550151 is a deficient number, since it is larger than the sum of its proper divisors (1).

131344422550151 is an equidigital number, since it uses as much as digits as its factorization.

131344422550151 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 288000, while the sum is 41.

Adding to 131344422550151 its reverse (151055224443131), we get a palindrome (282399646993282).

The spelling of 131344422550151 in words is "one hundred thirty-one trillion, three hundred forty-four billion, four hundred twenty-two million, five hundred fifty thousand, one hundred fifty-one".