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199906954433 is a prime number
BaseRepresentation
bin1011101000101101100…
…0100000110011000001
3201002222212012110222022
42322023120200303001
511233402140020213
6231500323451225
720304611035064
oct2721330406301
9632885173868
10199906954433
117786429a73a
12328b0526b15
1315b0ac71c24
1499659c08db
1553001e5308
hex2e8b620cc1

199906954433 has 2 divisors, whose sum is σ = 199906954434. Its totient is φ = 199906954432.

The previous prime is 199906954427. The next prime is 199906954441. The reversal of 199906954433 is 334459609991.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 105892969744 + 94013984689 = 325412^2 + 306617^2 .

It is a cyclic number.

It is not a de Polignac number, because 199906954433 - 24 = 199906954417 is a prime.

It is not a weakly prime, because it can be changed into another prime (199906954633) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 99953477216 + 99953477217.

It is an arithmetic number, because the mean of its divisors is an integer number (99953477217).

Almost surely, 2199906954433 is an apocalyptic number.

It is an amenable number.

199906954433 is a deficient number, since it is larger than the sum of its proper divisors (1).

199906954433 is an equidigital number, since it uses as much as digits as its factorization.

199906954433 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 28343520, while the sum is 62.

It can be divided in two parts, 1999069 and 54433, that added together give a palindrome (2053502).

The spelling of 199906954433 in words is "one hundred ninety-nine billion, nine hundred six million, nine hundred fifty-four thousand, four hundred thirty-three".