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136425351142 = 268212675571
BaseRepresentation
bin111111100001110010…
…1011010011111100110
3111001010201100220101001
41333003211122133212
54213344342214032
6142401204423514
712566514231343
oct1770345323746
9431121326331
10136425351142
1152948610593
12225346a7b9a
13cb32119b62
14686299b7ca
153836ebb3e7
hex1fc395a7e6

136425351142 has 4 divisors (see below), whose sum is σ = 204638026716. Its totient is φ = 68212675570.

The previous prime is 136425351121. The next prime is 136425351193. The reversal of 136425351142 is 241153524631.

It is a semiprime because it is the product of two primes.

It is a super-3 number, since 3×1364253511423 (a number of 34 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a congruent number.

It is an unprimeable number.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 34106337784 + ... + 34106337787.

It is an arithmetic number, because the mean of its divisors is an integer number (51159506679).

Almost surely, 2136425351142 is an apocalyptic number.

136425351142 is a deficient number, since it is larger than the sum of its proper divisors (68212675574).

136425351142 is an equidigital number, since it uses as much as digits as its factorization.

136425351142 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 68212675573.

The product of its digits is 86400, while the sum is 37.

Adding to 136425351142 its reverse (241153524631), we get a palindrome (377578875773).

The spelling of 136425351142 in words is "one hundred thirty-six billion, four hundred twenty-five million, three hundred fifty-one thousand, one hundred forty-two".

Divisors: 1 2 68212675571 136425351142