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43424733132113 is a prime number
BaseRepresentation
bin10011101111110100110111…
…10010101111010101010001
312200202100211022111221122112
421313322123302233111101
521142432323140211423
6232205014052515105
712101222240255534
oct1167723362572521
9180670738457575
1043424733132113
11129223648553a3
124a5400731ba95
131b2cc1c8a1bc8
14aa1aa82d0a1b
1550489c3b4a78
hex277e9bcaf551

43424733132113 has 2 divisors, whose sum is σ = 43424733132114. Its totient is φ = 43424733132112.

The previous prime is 43424733132089. The next prime is 43424733132167. The reversal of 43424733132113 is 31123133742434.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 41391633578689 + 2033099553424 = 6433633^2 + 1425868^2 .

It is a cyclic number.

It is not a de Polignac number, because 43424733132113 - 230 = 43423659390289 is a prime.

It is a super-2 number, since 2×434247331321132 (a number of 28 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (43424793132113) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 21712366566056 + 21712366566057.

It is an arithmetic number, because the mean of its divisors is an integer number (21712366566057).

Almost surely, 243424733132113 is an apocalyptic number.

It is an amenable number.

43424733132113 is a deficient number, since it is larger than the sum of its proper divisors (1).

43424733132113 is an equidigital number, since it uses as much as digits as its factorization.

43424733132113 is an evil number, because the sum of its binary digits is even.

The product of its digits is 435456, while the sum is 41.

Adding to 43424733132113 its reverse (31123133742434), we get a palindrome (74547866874547).

The spelling of 43424733132113 in words is "forty-three trillion, four hundred twenty-four billion, seven hundred thirty-three million, one hundred thirty-two thousand, one hundred thirteen".