Base | Representation |
---|---|
bin | 111111001011001100001… |
… | …110110101100000101111 |
3 | 120101000202200112120112221 |
4 | 333023030032311200233 |
5 | 1032112032403111343 |
6 | 13122215143540211 |
7 | 625436352060241 |
oct | 77131416654057 |
9 | 16330680476487 |
10 | 4341343410223 |
11 | 14241761a3996 |
12 | 5a146b15a667 |
13 | 256505103691 |
14 | 11019d230091 |
15 | 77ddd0685ed |
hex | 3f2cc3b582f |
4341343410223 has 2 divisors, whose sum is σ = 4341343410224. Its totient is φ = 4341343410222.
The previous prime is 4341343410179. The next prime is 4341343410259. The reversal of 4341343410223 is 3220143431434.
4341343410223 is digitally balanced in base 3, because in such base it contains all the possibile digits an equal number of times.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 4341343410223 - 213 = 4341343402031 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (4341343420223) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2170671705111 + 2170671705112.
It is an arithmetic number, because the mean of its divisors is an integer number (2170671705112).
Almost surely, 24341343410223 is an apocalyptic number.
4341343410223 is a deficient number, since it is larger than the sum of its proper divisors (1).
4341343410223 is an equidigital number, since it uses as much as digits as its factorization.
4341343410223 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 82944, while the sum is 34.
Adding to 4341343410223 its reverse (3220143431434), we get a palindrome (7561486841657).
The spelling of 4341343410223 in words is "four trillion, three hundred forty-one billion, three hundred forty-three million, four hundred ten thousand, two hundred twenty-three".
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