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142260211307 is a prime number
BaseRepresentation
bin1000010001111101011…
…1101000111001101011
3111121012100200111210002
42010133113220321223
54312322103230212
6145204202013215
713164224562434
oct2043727507153
9447170614702
10142260211307
11553721a9375
12236a27a720b
1310551c15836
146c578a048b
153a79383ac2
hex211f5e8e6b

142260211307 has 2 divisors, whose sum is σ = 142260211308. Its totient is φ = 142260211306.

The previous prime is 142260211231. The next prime is 142260211333. The reversal of 142260211307 is 703112062241.

It is a strong prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-142260211307 is a prime.

It is a super-2 number, since 2×1422602113072 (a number of 23 digits) contains 22 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 142260211307.

It is not a weakly prime, because it can be changed into another prime (142260211367) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 71130105653 + 71130105654.

It is an arithmetic number, because the mean of its divisors is an integer number (71130105654).

Almost surely, 2142260211307 is an apocalyptic number.

142260211307 is a deficient number, since it is larger than the sum of its proper divisors (1).

142260211307 is an equidigital number, since it uses as much as digits as its factorization.

142260211307 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 4032, while the sum is 29.

Adding to 142260211307 its reverse (703112062241), we get a palindrome (845372273548).

The spelling of 142260211307 in words is "one hundred forty-two billion, two hundred sixty million, two hundred eleven thousand, three hundred seven".