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202113112153747 is a prime number
BaseRepresentation
bin101101111101001000100000…
…010000010001111010010011
3222111121211021021221200021011
4231331020200100101322103
5202442410123202404442
61553505252032311351
760400122222260554
oct5575104020217223
9874554237850234
10202113112153747
1159443764807433
121a802a603a2557
1388a1252a12132
1437ca480b3632b
15185765c4ae117
hexb7d220411e93

202113112153747 has 2 divisors, whose sum is σ = 202113112153748. Its totient is φ = 202113112153746.

The previous prime is 202113112153699. The next prime is 202113112153759. The reversal of 202113112153747 is 747351211311202.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 202113112153747 - 215 = 202113112120979 is a prime.

It is a super-2 number, since 2×2021131121537472 (a number of 29 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (202113112154747) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 101056556076873 + 101056556076874.

It is an arithmetic number, because the mean of its divisors is an integer number (101056556076874).

Almost surely, 2202113112153747 is an apocalyptic number.

202113112153747 is a deficient number, since it is larger than the sum of its proper divisors (1).

202113112153747 is an equidigital number, since it uses as much as digits as its factorization.

202113112153747 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 70560, while the sum is 40.

Adding to 202113112153747 its reverse (747351211311202), we get a palindrome (949464323464949).

The spelling of 202113112153747 in words is "two hundred two trillion, one hundred thirteen billion, one hundred twelve million, one hundred fifty-three thousand, seven hundred forty-seven".