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1130214434143 = 15532179333989
BaseRepresentation
bin10000011100100110000…
…001111010010101011111
311000001021120212001110011
4100130212001322111133
5122004134343343033
62223114025424051
7144440524115554
oct20344601722537
94001246761404
101130214434143
113a6359646623
1216306263b027
138276a70c582
143c9ba1b0a2b
151e5ed335ccd
hex1072607a55f

1130214434143 has 8 divisors (see below), whose sum is σ = 1131464602800. Its totient is φ = 1128964940928.

The previous prime is 1130214434087. The next prime is 1130214434159. The reversal of 1130214434143 is 3414344120311.

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

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-1130214434143 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 (1130214434443) 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, 3216993 + ... + 3550981.

It is an arithmetic number, because the mean of its divisors is an integer number (141433075350).

Almost surely, 21130214434143 is an apocalyptic number.

1130214434143 is a deficient number, since it is larger than the sum of its proper divisors (1250168657).

1130214434143 is a wasteful number, since it uses less digits than its factorization.

1130214434143 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 337721.

The product of its (nonzero) digits is 13824, while the sum is 31.

Adding to 1130214434143 its reverse (3414344120311), we get a palindrome (4544558554454).

The spelling of 1130214434143 in words is "one trillion, one hundred thirty billion, two hundred fourteen million, four hundred thirty-four thousand, one hundred forty-three".

Divisors: 1 1553 2179 333989 3383987 518684917 727762031 1130214434143