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162454013537 is a prime number
BaseRepresentation
bin1001011101001100000…
…0111000011001100001
3120112022200212221201012
42113103000320121201
510130201211413122
6202344053150305
714510522221424
oct2272300703141
9515280787635
10162454013537
1162995067922
122759962b395
131241c7b773a
147c117d39bb
15435c1327e2
hex25d3038661

162454013537 has 2 divisors, whose sum is σ = 162454013538. Its totient is φ = 162454013536.

The previous prime is 162454013507. The next prime is 162454013539. The reversal of 162454013537 is 735310454261.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 162319546321 + 134467216 = 402889^2 + 11596^2 .

It is a cyclic number.

It is not a de Polignac number, because 162454013537 - 216 = 162453948001 is a prime.

Together with 162454013539, it forms a pair of twin primes.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 162454013494 and 162454013503.

It is not a weakly prime, because it can be changed into another prime (162454013539) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 81227006768 + 81227006769.

It is an arithmetic number, because the mean of its divisors is an integer number (81227006769).

Almost surely, 2162454013537 is an apocalyptic number.

It is an amenable number.

162454013537 is a deficient number, since it is larger than the sum of its proper divisors (1).

162454013537 is an equidigital number, since it uses as much as digits as its factorization.

162454013537 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 302400, while the sum is 41.

Adding to 162454013537 its reverse (735310454261), we get a palindrome (897764467798).

The spelling of 162454013537 in words is "one hundred sixty-two billion, four hundred fifty-four million, thirteen thousand, five hundred thirty-seven".