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201601882492273 is a prime number
BaseRepresentation
bin101101110101101100011000…
…100101111000110101110001
3222102210221200211011001110201
4231311230120211320311301
5202411021123404223043
61552434343134355201
760315153404662324
oct5565543045706561
9872727624131421
10201601882492273
1159266972a0a07a
121a73b96665bb01
138864c8c1c8873
1437ad82433c7bb
1518491daa73c4d
hexb75b18978d71

201601882492273 has 2 divisors, whose sum is σ = 201601882492274. Its totient is φ = 201601882492272.

The previous prime is 201601882492187. The next prime is 201601882492277. The reversal of 201601882492273 is 372294288106102.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 171493089318369 + 30108793173904 = 13095537^2 + 5487148^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-201601882492273 is a prime.

It is a super-3 number, since 3×2016018824922733 (a number of 44 digits) contains 333 as substring.

It is not a weakly prime, because it can be changed into another prime (201601882492277) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 100800941246136 + 100800941246137.

It is an arithmetic number, because the mean of its divisors is an integer number (100800941246137).

Almost surely, 2201601882492273 is an apocalyptic number.

It is an amenable number.

201601882492273 is a deficient number, since it is larger than the sum of its proper divisors (1).

201601882492273 is an equidigital number, since it uses as much as digits as its factorization.

201601882492273 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 4644864, while the sum is 55.

The spelling of 201601882492273 in words is "two hundred one trillion, six hundred one billion, eight hundred eighty-two million, four hundred ninety-two thousand, two hundred seventy-three".