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1320301433100 = 223521315114531543
BaseRepresentation
bin10011001101101000000…
…110010001000100001100
311200012221002002002221210
4103031220012101010030
5133112434131324400
62450312121454420
7164250205004106
oct23155006210414
94605832062853
101320301433100
11469a33814381
12193a72341410
1397672082250
1447c8d8d8576
1524526384950
hex1336819110c

1320301433100 has 288 divisors, whose sum is σ = 4146704149504. Its totient is φ = 322413696000.

The previous prime is 1320301433093. The next prime is 1320301433119. The reversal of 1320301433100 is 13341030231.

It is a super-2 number, since 2×13203014331002 (a number of 25 digits) contains 22 as substring.

It is an unprimeable number.

It is a polite number, since it can be written in 95 ways as a sum of consecutive naturals, for example, 855670929 + ... + 855672471.

Almost surely, 21320301433100 is an apocalyptic number.

1320301433100 is a gapful number since it is divisible by the number (10) formed by its first and last digit.

It is an amenable number.

It is a practical number, because each smaller number is the sum of distinct divisors of 1320301433100, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (2073352074752).

1320301433100 is an abundant number, since it is smaller than the sum of its proper divisors (2826402716404).

It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.

1320301433100 is a wasteful number, since it uses less digits than its factorization.

1320301433100 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 3177 (or 3170 counting only the distinct ones).

The product of its (nonzero) digits is 648, while the sum is 21.

Adding to 1320301433100 its reverse (13341030231), we get a palindrome (1333642463331).

The spelling of 1320301433100 in words is "one trillion, three hundred twenty billion, three hundred one million, four hundred thirty-three thousand, one hundred".