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13353124341553 is a prime number
BaseRepresentation
bin1100001001010000010000…
…1110111010011100110001
31202021112202012020201112021
43002110010032322130301
53222234144312412203
644222201324431441
72545505604105223
oct302240416723461
952245665221467
1013353124341553
1142890346370a7
1215b7b123a8581
1375b26a925349
1434241a7ca413
1518252ab415bd
hexc25043ba731

13353124341553 has 2 divisors, whose sum is σ = 13353124341554. Its totient is φ = 13353124341552.

The previous prime is 13353124341527. The next prime is 13353124341607. The reversal of 13353124341553 is 35514342135331.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 7410416395264 + 5942707946289 = 2722208^2 + 2437767^2 .

It is a cyclic number.

It is not a de Polignac number, because 13353124341553 - 29 = 13353124341041 is a prime.

It is not a weakly prime, because it can be changed into another prime (13353127341553) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6676562170776 + 6676562170777.

It is an arithmetic number, because the mean of its divisors is an integer number (6676562170777).

Almost surely, 213353124341553 is an apocalyptic number.

It is an amenable number.

13353124341553 is a deficient number, since it is larger than the sum of its proper divisors (1).

13353124341553 is an equidigital number, since it uses as much as digits as its factorization.

13353124341553 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 972000, while the sum is 43.

Adding to 13353124341553 its reverse (35514342135331), we get a palindrome (48867466476884).

The spelling of 13353124341553 in words is "thirteen trillion, three hundred fifty-three billion, one hundred twenty-four million, three hundred forty-one thousand, five hundred fifty-three".