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1103300134151 is a prime number
BaseRepresentation
bin10000000011100001110…
…100000000010100000111
310220110210212211221201112
4100003201310000110013
5121034024313243101
62202503222515235
7142465543255625
oct20034164002407
93813725757645
101103300134151
113959a8187696
121599b0b95b1b
138006b706687
143b5858ba515
151da7557d6bb
hex100e1d00507

1103300134151 has 2 divisors, whose sum is σ = 1103300134152. Its totient is φ = 1103300134150.

The previous prime is 1103300134121. The next prime is 1103300134207. The reversal of 1103300134151 is 1514310033011.

It is a weak prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-1103300134151 is a prime.

It is a super-2 number, since 2×11033001341512 (a number of 25 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1103300134121) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (13) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 551650067075 + 551650067076.

It is an arithmetic number, because the mean of its divisors is an integer number (551650067076).

Almost surely, 21103300134151 is an apocalyptic number.

1103300134151 is a deficient number, since it is larger than the sum of its proper divisors (1).

1103300134151 is an equidigital number, since it uses as much as digits as its factorization.

1103300134151 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 540, while the sum is 23.

Adding to 1103300134151 its reverse (1514310033011), we get a palindrome (2617610167162).

The spelling of 1103300134151 in words is "one trillion, one hundred three billion, three hundred million, one hundred thirty-four thousand, one hundred fifty-one".