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643653153125441 is a prime number
BaseRepresentation
bin1001001001011001100010110…
…1100011001111010001000001
310010101222122021220221001112122
42102112120231203033101001
51133331103124200003231
610200534102042334025
7252401256106462304
oct22226305543172101
93111878256831478
10643653153125441
111770a6850a33128
12602343627ba315
132181c30263a3a4
14b4d2d9bcbab3b
154e62d67a8077b
hex249662d8cf441

643653153125441 has 2 divisors, whose sum is σ = 643653153125442. Its totient is φ = 643653153125440.

The previous prime is 643653153125377. The next prime is 643653153125471. The reversal of 643653153125441 is 144521351356346.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 640616550643216 + 3036602482225 = 25310404^2 + 1742585^2 .

It is a cyclic number.

It is not a de Polignac number, because 643653153125441 - 26 = 643653153125377 is a prime.

It is not a weakly prime, because it can be changed into another prime (643653153125471) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 321826576562720 + 321826576562721.

It is an arithmetic number, because the mean of its divisors is an integer number (321826576562721).

Almost surely, 2643653153125441 is an apocalyptic number.

It is an amenable number.

643653153125441 is a deficient number, since it is larger than the sum of its proper divisors (1).

643653153125441 is an equidigital number, since it uses as much as digits as its factorization.

643653153125441 is an evil number, because the sum of its binary digits is even.

The product of its digits is 15552000, while the sum is 53.

The spelling of 643653153125441 in words is "six hundred forty-three trillion, six hundred fifty-three billion, one hundred fifty-three million, one hundred twenty-five thousand, four hundred forty-one".