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1115111312141 is a prime number
BaseRepresentation
bin10000001110100001110…
…100000100101100001101
310221121022000122221222012
4100032201310010230031
5121232221443442031
62212135225231005
7143364332522111
oct20164164045415
93847260587865
101115111312141
1139aa092a0512
12160148632465
13812016c89a8
143bd863c5741
151e017451b2b
hex103a1d04b0d

1115111312141 has 2 divisors, whose sum is σ = 1115111312142. Its totient is φ = 1115111312140.

The previous prime is 1115111312137. The next prime is 1115111312261. The reversal of 1115111312141 is 1412131115111.

It is a m-pointer prime, because the next prime (1115111312261) can be obtained adding 1115111312141 to its product of digits (120).

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 588404055625 + 526707256516 = 767075^2 + 725746^2 .

It is a cyclic number.

It is not a de Polignac number, because 1115111312141 - 22 = 1115111312137 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1115111312101) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 557555656070 + 557555656071.

It is an arithmetic number, because the mean of its divisors is an integer number (557555656071).

Almost surely, 21115111312141 is an apocalyptic number.

It is an amenable number.

1115111312141 is a deficient number, since it is larger than the sum of its proper divisors (1).

1115111312141 is an equidigital number, since it uses as much as digits as its factorization.

1115111312141 is an evil number, because the sum of its binary digits is even.

The product of its digits is 120, while the sum is 23.

Adding to 1115111312141 its reverse (1412131115111), we get a palindrome (2527242427252).

The spelling of 1115111312141 in words is "one trillion, one hundred fifteen billion, one hundred eleven million, three hundred twelve thousand, one hundred forty-one".