Base | Representation |
---|---|
bin | 1000010101001000101… |
… | …0101000110110000001 |
3 | 111200101202022021102211 |
4 | 2011102022220312001 |
5 | 4321043241104332 |
6 | 145424524142121 |
7 | 13224313451644 |
oct | 2052212506601 |
9 | 450352267384 |
10 | 143112441217 |
11 | 5576a273309 |
12 | 238a0097941 |
13 | 10659654aa5 |
14 | 6cd8b43a5b |
15 | 3ac90c6447 |
hex | 21522a8d81 |
143112441217 has 2 divisors, whose sum is σ = 143112441218. Its totient is φ = 143112441216.
The previous prime is 143112441209. The next prime is 143112441259. The reversal of 143112441217 is 712144211341.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 87684685456 + 55427755761 = 296116^2 + 235431^2 .
It is a cyclic number.
It is not a de Polignac number, because 143112441217 - 23 = 143112441209 is a prime.
It is a super-3 number, since 3×1431124412173 (a number of 34 digits) contains 333 as substring.
It is not a weakly prime, because it can be changed into another prime (143112441517) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 71556220608 + 71556220609.
It is an arithmetic number, because the mean of its divisors is an integer number (71556220609).
Almost surely, 2143112441217 is an apocalyptic number.
It is an amenable number.
143112441217 is a deficient number, since it is larger than the sum of its proper divisors (1).
143112441217 is an equidigital number, since it uses as much as digits as its factorization.
143112441217 is an evil number, because the sum of its binary digits is even.
The product of its digits is 5376, while the sum is 31.
Adding to 143112441217 its reverse (712144211341), we get a palindrome (855256652558).
The spelling of 143112441217 in words is "one hundred forty-three billion, one hundred twelve million, four hundred forty-one thousand, two hundred seventeen".
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