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13501332553433 is a prime number
BaseRepresentation
bin1100010001111000011000…
…1000010111111011011001
31202210201022002010222011212
43010132012020113323121
53232201212113202213
644414232045032505
72562303415113344
oct304360610277331
952721262128155
1013501332553433
11433597927a31a
121620794b55735
1376c229bcb43a
1434967a20145b
1518630228a9a8
hexc4786217ed9

13501332553433 has 2 divisors, whose sum is σ = 13501332553434. Its totient is φ = 13501332553432.

The previous prime is 13501332553421. The next prime is 13501332553519. The reversal of 13501332553433 is 33435523310531.

13501332553433 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 11275277505424 + 2226055048009 = 3357868^2 + 1491997^2 .

It is a cyclic number.

It is not a de Polignac number, because 13501332553433 - 216 = 13501332487897 is a prime.

It is not a weakly prime, because it can be changed into another prime (13501332553133) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6750666276716 + 6750666276717.

It is an arithmetic number, because the mean of its divisors is an integer number (6750666276717).

Almost surely, 213501332553433 is an apocalyptic number.

It is an amenable number.

13501332553433 is a deficient number, since it is larger than the sum of its proper divisors (1).

13501332553433 is an equidigital number, since it uses as much as digits as its factorization.

13501332553433 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 729000, while the sum is 41.

Adding to 13501332553433 its reverse (33435523310531), we get a palindrome (46936855863964).

The spelling of 13501332553433 in words is "thirteen trillion, five hundred one billion, three hundred thirty-two million, five hundred fifty-three thousand, four hundred thirty-three".