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1533534040041 = 3737002438539
BaseRepresentation
bin10110010100001101110…
…000000110111111101001
312102121022121211212122120
4112110031300012333221
5200111134203240131
63132255032341453
7215536255160103
oct26241560067751
95377277755576
101533534040041
11541405a93a73
12209261544889
13b17c4a05b53
145431b1b4573
1529d563a6496
hex1650dc06fe9

1533534040041 has 8 divisors (see below), whose sum is σ = 2072721807840. Its totient is φ = 1008351149472.

The previous prime is 1533534040001. The next prime is 1533534040079. The reversal of 1533534040041 is 1400404353351.

It is a sphenic number, since it is the product of 3 distinct primes.

It is not a de Polignac number, because 1533534040041 - 27 = 1533534039913 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 1533534039984 and 1533534040011.

It is not an unprimeable number, because it can be changed into a prime (1533534040001) by changing a digit.

It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 3501219051 + ... + 3501219488.

It is an arithmetic number, because the mean of its divisors is an integer number (259090225980).

Almost surely, 21533534040041 is an apocalyptic number.

It is an amenable number.

1533534040041 is a deficient number, since it is larger than the sum of its proper divisors (539187767799).

1533534040041 is an equidigital number, since it uses as much as digits as its factorization.

1533534040041 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 7002438615.

The product of its (nonzero) digits is 43200, while the sum is 33.

Adding to 1533534040041 its reverse (1400404353351), we get a palindrome (2933938393392).

The spelling of 1533534040041 in words is "one trillion, five hundred thirty-three billion, five hundred thirty-four million, forty thousand, forty-one".

Divisors: 1 3 73 219 7002438539 21007315617 511178013347 1533534040041