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641214173437 is a prime number
BaseRepresentation
bin10010101010010110101…
…01111101000011111101
32021022002022120222211021
421111023111331003331
541001201312022222
61210323021024141
764216612355035
oct11251325750375
92238068528737
10641214173437
11227a34759062
12a43313b5051
134860a427596
142306bbd08c5
1511a2d23eac7
hex954b57d0fd

641214173437 has 2 divisors, whose sum is σ = 641214173438. Its totient is φ = 641214173436.

The previous prime is 641214173393. The next prime is 641214173441. The reversal of 641214173437 is 734371412146.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 480628839076 + 160585334361 = 693274^2 + 400731^2 .

It is a cyclic number.

It is not a de Polignac number, because 641214173437 - 215 = 641214140669 is a prime.

It is a super-2 number, since 2×6412141734372 (a number of 24 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 641214173393 and 641214173402.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (641214174437) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 320607086718 + 320607086719.

It is an arithmetic number, because the mean of its divisors is an integer number (320607086719).

Almost surely, 2641214173437 is an apocalyptic number.

It is an amenable number.

641214173437 is a deficient number, since it is larger than the sum of its proper divisors (1).

641214173437 is an equidigital number, since it uses as much as digits as its factorization.

641214173437 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 338688, while the sum is 43.

The spelling of 641214173437 in words is "six hundred forty-one billion, two hundred fourteen million, one hundred seventy-three thousand, four hundred thirty-seven".