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43330425151 = 133333109627
BaseRepresentation
bin101000010110101100…
…011101000100111111
311010211202211122221011
4220112230131010333
51202220042101101
631523345220051
73062524352056
oct502654350477
9133752748834
1043330425151
1117415993822
128493331627
134117058230
142150a2739d
1511d9099751
hexa16b1d13f

43330425151 has 4 divisors (see below), whose sum is σ = 46663534792. Its totient is φ = 39997315512.

The previous prime is 43330425143. The next prime is 43330425191. The reversal of 43330425151 is 15152403334.

It is a semiprime because it is the product of two primes.

It is a cyclic number.

It is not a de Polignac number, because 43330425151 - 23 = 43330425143 is a prime.

It is a Duffinian number.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (43330425191) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1666554801 + ... + 1666554826.

It is an arithmetic number, because the mean of its divisors is an integer number (11665883698).

Almost surely, 243330425151 is an apocalyptic number.

43330425151 is a deficient number, since it is larger than the sum of its proper divisors (3333109641).

43330425151 is a wasteful number, since it uses less digits than its factorization.

43330425151 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 3333109640.

The product of its (nonzero) digits is 21600, while the sum is 31.

Adding to 43330425151 its reverse (15152403334), we get a palindrome (58482828485).

The spelling of 43330425151 in words is "forty-three billion, three hundred thirty million, four hundred twenty-five thousand, one hundred fifty-one".

Divisors: 1 13 3333109627 43330425151