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1420152012553 is a prime number
BaseRepresentation
bin10100101010100111101…
…001111111111100001001
312000202122211120002202011
4110222213221333330021
5141231432403400203
63004224150135521
7204413455635151
oct24524751777411
95022584502664
101420152012553
114a831287361a
121ab29a0515a1
13a3bc5936761
144ca42b81761
1526e1c40676d
hex14aa7a7ff09

1420152012553 has 2 divisors, whose sum is σ = 1420152012554. Its totient is φ = 1420152012552.

The previous prime is 1420152012463. The next prime is 1420152012557. The reversal of 1420152012553 is 3552102510241.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 1384639363849 + 35512648704 = 1176707^2 + 188448^2 .

It is an emirp because it is prime and its reverse (3552102510241) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 1420152012553 - 217 = 1420151881481 is a prime.

It is not a weakly prime, because it can be changed into another prime (1420152012557) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 710076006276 + 710076006277.

It is an arithmetic number, because the mean of its divisors is an integer number (710076006277).

Almost surely, 21420152012553 is an apocalyptic number.

It is an amenable number.

1420152012553 is a deficient number, since it is larger than the sum of its proper divisors (1).

1420152012553 is an equidigital number, since it uses as much as digits as its factorization.

1420152012553 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 12000, while the sum is 31.

Adding to 1420152012553 its reverse (3552102510241), we get a palindrome (4972254522794).

The spelling of 1420152012553 in words is "one trillion, four hundred twenty billion, one hundred fifty-two million, twelve thousand, five hundred fifty-three".