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1100202020123 = 3473170611009
BaseRepresentation
bin10000000000101001001…
…001101000110100011011
310220011210222012110020112
4100000221021220310123
5121011203204120443
62201231551314535
7142326013604333
oct20005111506433
93804728173215
101100202020123
113946583a3778
1215928751ba4b
137c9968bbbcb
143b3702540c3
151d9435a8b18
hex10029268d1b

1100202020123 has 4 divisors (see below), whose sum is σ = 1103372631480. Its totient is φ = 1097031408768.

The previous prime is 1100202020107. The next prime is 1100202020131. The reversal of 1100202020123 is 3210202020011.

It is a happy number.

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 1100202020123 - 24 = 1100202020107 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 1100202020098 and 1100202020107.

It is not an unprimeable number, because it can be changed into a prime (1100202620123) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1585305158 + ... + 1585305851.

It is an arithmetic number, because the mean of its divisors is an integer number (275843157870).

Almost surely, 21100202020123 is an apocalyptic number.

1100202020123 is a deficient number, since it is larger than the sum of its proper divisors (3170611357).

1100202020123 is an equidigital number, since it uses as much as digits as its factorization.

1100202020123 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 3170611356.

The product of its (nonzero) digits is 48, while the sum is 14.

Adding to 1100202020123 its reverse (3210202020011), we get a palindrome (4310404040134).

The spelling of 1100202020123 in words is "one trillion, one hundred billion, two hundred two million, twenty thousand, one hundred twenty-three".

Divisors: 1 347 3170611009 1100202020123