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1310321554153 is a prime number
BaseRepresentation
bin10011000100010101010…
…000000011001011101001
311122021011120011201201121
4103010111100003023221
5132432014304213103
62441541534215241
7163444663304113
oct23042520031351
94567146151647
101310321554153
114657823a2637
12191b48054521
13967415bb331
14475c42d23b3
152414014d6bd
hex131154032e9

1310321554153 has 2 divisors, whose sum is σ = 1310321554154. Its totient is φ = 1310321554152.

The previous prime is 1310321554093. The next prime is 1310321554219. The reversal of 1310321554153 is 3514551230131.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 1264338327184 + 45983226969 = 1124428^2 + 214437^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-1310321554153 is a prime.

It is not a weakly prime, because it can be changed into another prime (1310321554253) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 655160777076 + 655160777077.

It is an arithmetic number, because the mean of its divisors is an integer number (655160777077).

Almost surely, 21310321554153 is an apocalyptic number.

It is an amenable number.

1310321554153 is a deficient number, since it is larger than the sum of its proper divisors (1).

1310321554153 is an equidigital number, since it uses as much as digits as its factorization.

1310321554153 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 27000, while the sum is 34.

Adding to 1310321554153 its reverse (3514551230131), we get a palindrome (4824872784284).

The spelling of 1310321554153 in words is "one trillion, three hundred ten billion, three hundred twenty-one million, five hundred fifty-four thousand, one hundred fifty-three".