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243121254133 is a prime number
BaseRepresentation
bin1110001001101100100…
…1111110111011110101
3212020112112112100201211
43202123021332323311
512440403020113013
6303404404121421
723364525364042
oct3423311767365
9766475470654
10243121254133
119411a67870a
123b150996271
1319c06c2b974
14baa501a1c9
1564cdea223d
hex389b27eef5

243121254133 has 2 divisors, whose sum is σ = 243121254134. Its totient is φ = 243121254132.

The previous prime is 243121254083. The next prime is 243121254149. The reversal of 243121254133 is 331452121342.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 180912414244 + 62208839889 = 425338^2 + 249417^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-243121254133 is a prime.

It is a super-2 number, since 2×2431212541332 (a number of 24 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 243121254095 and 243121254104.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (243121254533) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 121560627066 + 121560627067.

It is an arithmetic number, because the mean of its divisors is an integer number (121560627067).

Almost surely, 2243121254133 is an apocalyptic number.

It is an amenable number.

243121254133 is a deficient number, since it is larger than the sum of its proper divisors (1).

243121254133 is an equidigital number, since it uses as much as digits as its factorization.

243121254133 is an evil number, because the sum of its binary digits is even.

The product of its digits is 17280, while the sum is 31.

Adding to 243121254133 its reverse (331452121342), we get a palindrome (574573375475).

The spelling of 243121254133 in words is "two hundred forty-three billion, one hundred twenty-one million, two hundred fifty-four thousand, one hundred thirty-three".