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10203044213147 is a prime number
BaseRepresentation
bin1001010001111001010011…
…0100110111000110011011
31100010101211200201222200212
42110132110310313012123
52314131313304310042
633411114103121335
72102100642600302
oct224362464670633
940111750658625
1010203044213147
1132840a5513655
121189502a4324b
135901b206c346
14273b89645439
1512a610791482
hex94794d3719b

10203044213147 has 2 divisors, whose sum is σ = 10203044213148. Its totient is φ = 10203044213146.

The previous prime is 10203044213093. The next prime is 10203044213231. The reversal of 10203044213147 is 74131244030201.

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 10203044213147 - 26 = 10203044213083 is a prime.

It is a super-2 number, since 2×102030442131472 (a number of 27 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (10203044210147) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5101522106573 + 5101522106574.

It is an arithmetic number, because the mean of its divisors is an integer number (5101522106574).

Almost surely, 210203044213147 is an apocalyptic number.

10203044213147 is a deficient number, since it is larger than the sum of its proper divisors (1).

10203044213147 is an equidigital number, since it uses as much as digits as its factorization.

10203044213147 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 16128, while the sum is 32.

Adding to 10203044213147 its reverse (74131244030201), we get a palindrome (84334288243348).

The spelling of 10203044213147 in words is "ten trillion, two hundred three billion, forty-four million, two hundred thirteen thousand, one hundred forty-seven".