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11411312103437 is a prime number
BaseRepresentation
bin1010011000001110011100…
…1101000101000000001101
31111101220120221221102222012
42212003213031011000031
52443430331344302222
640134141233231005
72255303644316621
oct246034715050015
944356527842865
1011411312103437
1136aa568210024
121343709402465
1364a10b53490c
142b644bc4d181
1514bc7b2421e2
hexa60e734500d

11411312103437 has 2 divisors, whose sum is σ = 11411312103438. Its totient is φ = 11411312103436.

The previous prime is 11411312103413. The next prime is 11411312103469. The reversal of 11411312103437 is 73430121311411.

It is an a-pointer prime, because the next prime (11411312103469) can be obtained adding 11411312103437 to its sum of digits (32).

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 11400745497001 + 10566606436 = 3376499^2 + 102794^2 .

It is a cyclic number.

It is not a de Polignac number, because 11411312103437 - 28 = 11411312103181 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (11411312103407) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5705656051718 + 5705656051719.

It is an arithmetic number, because the mean of its divisors is an integer number (5705656051719).

Almost surely, 211411312103437 is an apocalyptic number.

It is an amenable number.

11411312103437 is a deficient number, since it is larger than the sum of its proper divisors (1).

11411312103437 is an equidigital number, since it uses as much as digits as its factorization.

11411312103437 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 6048, while the sum is 32.

Adding to 11411312103437 its reverse (73430121311411), we get a palindrome (84841433414848).

The spelling of 11411312103437 in words is "eleven trillion, four hundred eleven billion, three hundred twelve million, one hundred three thousand, four hundred thirty-seven".