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143123600556049 is a prime number
BaseRepresentation
bin100000100010101110001111…
…010111101100000000010001
3200202202111000210202121220111
4200202232033113230000101
5122224414113220243144
61224222024023241321
742101224146654421
oct4042561727540021
9622674023677814
10143123600556049
11416704183200a3
12140763902a9241
1361b2661370ba9
14274b30d7c2481
151182e8ed57534
hex822b8f5ec011

143123600556049 has 2 divisors, whose sum is σ = 143123600556050. Its totient is φ = 143123600556048.

The previous prime is 143123600556019. The next prime is 143123600556173. The reversal of 143123600556049 is 940655006321341.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 110152749455424 + 32970851100625 = 10495368^2 + 5742025^2 .

It is a cyclic number.

It is not a de Polignac number, because 143123600556049 - 27 = 143123600555921 is a prime.

It is a super-2 number, since 2×1431236005560492 (a number of 29 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 143123600555993 and 143123600556011.

It is not a weakly prime, because it can be changed into another prime (143123600556019) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 71561800278024 + 71561800278025.

It is an arithmetic number, because the mean of its divisors is an integer number (71561800278025).

Almost surely, 2143123600556049 is an apocalyptic number.

It is an amenable number.

143123600556049 is a deficient number, since it is larger than the sum of its proper divisors (1).

143123600556049 is an equidigital number, since it uses as much as digits as its factorization.

143123600556049 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 2332800, while the sum is 49.

The spelling of 143123600556049 in words is "one hundred forty-three trillion, one hundred twenty-three billion, six hundred million, five hundred fifty-six thousand, forty-nine".