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145121554103 is a prime number
BaseRepresentation
bin1000011100100111101…
…0110010111010110111
3111212120202210210012012
42013021322302322313
54334202104212403
6150400134314435
713325150566631
oct2071172627267
9455522723165
10145121554103
1156600374589
1224160ab8a1b
13108b996cc96
1470498d1a51
153b95688ed8
hex21c9eb2eb7

145121554103 has 2 divisors, whose sum is σ = 145121554104. Its totient is φ = 145121554102.

The previous prime is 145121554099. The next prime is 145121554117. The reversal of 145121554103 is 301455121541.

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 145121554103 - 22 = 145121554099 is a prime.

It is a super-2 number, since 2×1451215541032 (a number of 23 digits) contains 22 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 145121554103.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (145121524103) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 72560777051 + 72560777052.

It is an arithmetic number, because the mean of its divisors is an integer number (72560777052).

Almost surely, 2145121554103 is an apocalyptic number.

145121554103 is a deficient number, since it is larger than the sum of its proper divisors (1).

145121554103 is an equidigital number, since it uses as much as digits as its factorization.

145121554103 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 12000, while the sum is 32.

Adding to 145121554103 its reverse (301455121541), we get a palindrome (446576675644).

The spelling of 145121554103 in words is "one hundred forty-five billion, one hundred twenty-one million, five hundred fifty-four thousand, one hundred three".