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115454433404347 is a prime number
BaseRepresentation
bin11010010000000101010100…
…011111011010010110111011
3120010210100001222021201120221
4122100011110133122112323
5110113101134422414342
61045315011241200511
733214210326501622
oct3220052437322673
9503710058251527
10115454433404347
1133872a4a847434
1210b47a14190137
134c563ca0a114b
14207204ad0c6b9
15d5337c508a67
hex6901547da5bb

115454433404347 has 2 divisors, whose sum is σ = 115454433404348. Its totient is φ = 115454433404346.

The previous prime is 115454433404327. The next prime is 115454433404357. The reversal of 115454433404347 is 743404334454511.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 115454433404347 - 215 = 115454433371579 is a prime.

It is a super-3 number, since 3×1154544334043473 (a number of 43 digits) contains 333 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 115454433404294 and 115454433404303.

It is not a weakly prime, because it can be changed into another prime (115454433404327) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 57727216702173 + 57727216702174.

It is an arithmetic number, because the mean of its divisors is an integer number (57727216702174).

Almost surely, 2115454433404347 is an apocalyptic number.

115454433404347 is a deficient number, since it is larger than the sum of its proper divisors (1).

115454433404347 is an equidigital number, since it uses as much as digits as its factorization.

115454433404347 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 19353600, while the sum is 52.

Adding to 115454433404347 its reverse (743404334454511), we get a palindrome (858858767858858).

The spelling of 115454433404347 in words is "one hundred fifteen trillion, four hundred fifty-four billion, four hundred thirty-three million, four hundred four thousand, three hundred forty-seven".