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1325203410433 = 75471422124337
BaseRepresentation
bin10011010010001100010…
…001110100111000000001
311200200120201000100200211
4103102030101310320001
5133203004033113213
62452442352114121
7164512530063160
oct23221421647001
94620521010624
101325203410433
11471019860511
121949bbb35941
1397c73810356
14481d69613d7
15247118c753d
hex1348c474e01

1325203410433 has 16 divisors (see below), whose sum is σ = 1517455997824. Its totient is φ = 1133685745920.

The previous prime is 1325203410431. The next prime is 1325203410469. The reversal of 1325203410433 is 3340143025231.

It is a happy number.

It is not a de Polignac number, because 1325203410433 - 21 = 1325203410431 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 1325203410395 and 1325203410404.

It is not an unprimeable number, because it can be changed into a prime (1325203410431) by changing a digit.

It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 54440041 + ... + 54464377.

It is an arithmetic number, because the mean of its divisors is an integer number (94840999864).

Almost surely, 21325203410433 is an apocalyptic number.

It is an amenable number.

1325203410433 is a deficient number, since it is larger than the sum of its proper divisors (192252587391).

1325203410433 is a wasteful number, since it uses less digits than its factorization.

1325203410433 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 39112.

The product of its (nonzero) digits is 25920, while the sum is 31.

Adding to 1325203410433 its reverse (3340143025231), we get a palindrome (4665346435664).

The spelling of 1325203410433 in words is "one trillion, three hundred twenty-five billion, two hundred three million, four hundred ten thousand, four hundred thirty-three".

Divisors: 1 7 547 3829 14221 24337 99547 170359 7778887 13312339 54452209 93186373 346096477 2422675339 189314772919 1325203410433