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131772018953 is a prime number
BaseRepresentation
bin111101010111000111…
…0010111000100001001
3110121010102022010122102
41322232032113010021
54124332114101303
6140311335403145
712343301452241
oct1725616270411
9417112263572
10131772018953
115097a962203
12216562234b5
13c570048863
14654097ab21
1536636e1288
hex1eae397109

131772018953 has 2 divisors, whose sum is σ = 131772018954. Its totient is φ = 131772018952.

The previous prime is 131772018901. The next prime is 131772018979. The reversal of 131772018953 is 359810277131.

It is a happy number.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 78129753289 + 53642265664 = 279517^2 + 231608^2 .

It is a cyclic number.

It is not a de Polignac number, because 131772018953 - 210 = 131772017929 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 131772018898 and 131772018907.

It is not a weakly prime, because it can be changed into another prime (131772018253) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65886009476 + 65886009477.

It is an arithmetic number, because the mean of its divisors is an integer number (65886009477).

Almost surely, 2131772018953 is an apocalyptic number.

It is an amenable number.

131772018953 is a deficient number, since it is larger than the sum of its proper divisors (1).

131772018953 is an equidigital number, since it uses as much as digits as its factorization.

131772018953 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 317520, while the sum is 47.

The spelling of 131772018953 in words is "one hundred thirty-one billion, seven hundred seventy-two million, eighteen thousand, nine hundred fifty-three".