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145101725113 is a prime number
BaseRepresentation
bin1000011100100010111…
…1001001110110111001
3111212112101111100221011
42013020233021312321
54334132020200423
6150354145313521
713324520212312
oct2071057116671
9455471440834
10145101725113
11565a0160794
12241563358a1
13108b5819657
14704700d609
153b93a6db0d
hex21c8bc9db9

145101725113 has 2 divisors, whose sum is σ = 145101725114. Its totient is φ = 145101725112.

The previous prime is 145101725101. The next prime is 145101725159. The reversal of 145101725113 is 311527101541.

It is a happy number.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 120998231104 + 24103494009 = 347848^2 + 155253^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-145101725113 is a prime.

It is a super-2 number, since 2×1451017251132 (a number of 23 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (145101725413) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 72550862556 + 72550862557.

It is an arithmetic number, because the mean of its divisors is an integer number (72550862557).

Almost surely, 2145101725113 is an apocalyptic number.

It is an amenable number.

145101725113 is a deficient number, since it is larger than the sum of its proper divisors (1).

145101725113 is an equidigital number, since it uses as much as digits as its factorization.

145101725113 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 4200, while the sum is 31.

Adding to 145101725113 its reverse (311527101541), we get a palindrome (456628826654).

The spelling of 145101725113 in words is "one hundred forty-five billion, one hundred one million, seven hundred twenty-five thousand, one hundred thirteen".