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1725132841 is a prime number
BaseRepresentation
bin110011011010011…
…0111000000101001
311110020010211100221
41212310313000221
512013113222331
6443103330041
760515246545
oct14664670051
94406124327
101725132841
11805879923
124018b0921
13216539083
14125188a25
15a16ba711
hex66d37029

1725132841 has 2 divisors, whose sum is σ = 1725132842. Its totient is φ = 1725132840.

The previous prime is 1725132809. The next prime is 1725132853. The reversal of 1725132841 is 1482315271.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 1449705625 + 275427216 = 38075^2 + 16596^2 .

It is a cyclic number.

It is not a de Polignac number, because 1725132841 - 25 = 1725132809 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 1725132797 and 1725132806.

It is not a weakly prime, because it can be changed into another prime (1725132041) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 862566420 + 862566421.

It is an arithmetic number, because the mean of its divisors is an integer number (862566421).

Almost surely, 21725132841 is an apocalyptic number.

It is an amenable number.

1725132841 is a deficient number, since it is larger than the sum of its proper divisors (1).

1725132841 is an equidigital number, since it uses as much as digits as its factorization.

1725132841 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 13440, while the sum is 34.

The square root of 1725132841 is about 41534.7185015139. The cubic root of 1725132841 is about 1199.3359383733.

The spelling of 1725132841 in words is "one billion, seven hundred twenty-five million, one hundred thirty-two thousand, eight hundred forty-one".