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4430224438151 is a prime number
BaseRepresentation
bin100000001110111110111…
…1101000011001110000111
3120200112011221212121120012
41000131331331003032013
51040041044424010101
613231114512500435
7635034043252451
oct100357575031607
916615157777505
104430224438151
111458936221776
125b673524a71b
132619cb1616b5
141145d16912d1
157a391062bbb
hex4077df43387

4430224438151 has 2 divisors, whose sum is σ = 4430224438152. Its totient is φ = 4430224438150.

The previous prime is 4430224438121. The next prime is 4430224438153. The reversal of 4430224438151 is 1518344220344.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 4430224438151 - 226 = 4430157329287 is a prime.

Together with 4430224438153, it forms a pair of twin primes.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 4430224438099 and 4430224438108.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (4430224438153) 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, 2215112219075 + 2215112219076.

It is an arithmetic number, because the mean of its divisors is an integer number (2215112219076).

Almost surely, 24430224438151 is an apocalyptic number.

4430224438151 is a deficient number, since it is larger than the sum of its proper divisors (1).

4430224438151 is an equidigital number, since it uses as much as digits as its factorization.

4430224438151 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 368640, while the sum is 41.

Adding to 4430224438151 its reverse (1518344220344), we get a palindrome (5948568658495).

The spelling of 4430224438151 in words is "four trillion, four hundred thirty billion, two hundred twenty-four million, four hundred thirty-eight thousand, one hundred fifty-one".