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120043010420189 is a prime number
BaseRepresentation
bin11011010010110110110000…
…111111100110110111011101
3120202000222222210110202120222
4123102312300333212313131
5111213241041131421224
61103150555255251125
734166551463550203
oct3322666077466735
9522028883422528
10120043010420189
1135281a5595a51a
1211569181379aa5
1351ca0130a164c
142190181397073
15dd28da58db5e
hex6d2db0fe6ddd

120043010420189 has 2 divisors, whose sum is σ = 120043010420190. Its totient is φ = 120043010420188.

The previous prime is 120043010420167. The next prime is 120043010420243. The reversal of 120043010420189 is 981024010340021.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 65714936925625 + 54328073494564 = 8106475^2 + 7370758^2 .

It is an emirp because it is prime and its reverse (981024010340021) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 120043010420189 - 228 = 120042741984733 is a prime.

It is a super-3 number, since 3×1200430104201893 (a number of 43 digits) contains 333 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (120043010420119) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 60021505210094 + 60021505210095.

It is an arithmetic number, because the mean of its divisors is an integer number (60021505210095).

Almost surely, 2120043010420189 is an apocalyptic number.

It is an amenable number.

120043010420189 is a deficient number, since it is larger than the sum of its proper divisors (1).

120043010420189 is an equidigital number, since it uses as much as digits as its factorization.

120043010420189 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 13824, while the sum is 35.

The spelling of 120043010420189 in words is "one hundred twenty trillion, forty-three billion, ten million, four hundred twenty thousand, one hundred eighty-nine".