Base | Representation |
---|---|
bin | 10100101111001111010… |
… | …111100000010101100011 |
3 | 12001020110120221120112211 |
4 | 110233033113200111203 |
5 | 141322112441023301 |
6 | 3010404334102551 |
7 | 204650430335134 |
oct | 24571727402543 |
9 | 5036416846484 |
10 | 1425113220451 |
11 | 4aa429295538 |
12 | 1b024364aa57 |
13 | a450673252b |
14 | 4cd93a25c8b |
15 | 2710cc49051 |
hex | 14bcf5e0563 |
1425113220451 has 2 divisors, whose sum is σ = 1425113220452. Its totient is φ = 1425113220450.
The previous prime is 1425113220397. The next prime is 1425113220469. The reversal of 1425113220451 is 1540223115241.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 1425113220451 - 215 = 1425113187683 is a prime.
It is a super-2 number, since 2×14251132204512 (a number of 25 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (1425113220551) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 712556610225 + 712556610226.
It is an arithmetic number, because the mean of its divisors is an integer number (712556610226).
Almost surely, 21425113220451 is an apocalyptic number.
1425113220451 is a deficient number, since it is larger than the sum of its proper divisors (1).
1425113220451 is an equidigital number, since it uses as much as digits as its factorization.
1425113220451 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 9600, while the sum is 31.
Adding to 1425113220451 its reverse (1540223115241), we get a palindrome (2965336335692).
The spelling of 1425113220451 in words is "one trillion, four hundred twenty-five billion, one hundred thirteen million, two hundred twenty thousand, four hundred fifty-one".
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