Okay, so I need to find the fraction equivalent of the repeating decimal 0.133113311331..., right? And then add the numerator and denominator after simplifying the fraction to its lowest terms. Let me think about how to approach this.

First, repeating decimals can usually be converted into fractions by using algebra. I remember that for a repeating decimal like 0.\overline{abc}, where "abc" repeats indefinitely, the method involves setting the decimal equal to a variable, multiplying both sides by 10^n (where n is the number of repeating digits) to shift the decimal point, and then subtracting the original equation to eliminate the repeating part.

In this case, the repeating part is "1331", which has four digits. So maybe I can use a similar method here. Let me try setting x = 0.133113311331...

Then, if I multiply both sides by 10,000 (since there are four repeating digits), I get 10,000x = 1331.13311331...

Now, if I subtract the original equation (x = 0.13311331...) from this new equation, the repeating parts should cancel out. Let's do that:

10,000x - x = 1331.13311331... - 0.13311331...
9,999x = 1331

So, solving for x gives x = 1331 / 9,999.

But wait, I need to check if this fraction can be simplified further. To do that, I need to find the greatest common divisor (GCD) of 1331 and 9,999.

Hmm, let's see. Let me factorize both numbers.

First, 1331. I recall that 11^3 is 1331 because 11*11=121, 121*11=1331. So 1331 = 11^3.

Now, 9,999. Let's divide this by 9 first. 9*1,111 = 9,999. Wait, 9*1,000=9,000, so 9,000 + 9*111=9,000 + 999=9,999. So 9,999 = 9*1,111.

But 1,111... Hmm, is 1,111 divisible by 11? Let's check. The divisibility rule for 11 is alternating sum of digits. For 1,111: (1 - 1 + 1 - 1) = 0, which is divisible by 11. So 1,111 ÷ 11 = 101. Let me verify: 11*101 = 1,111. Yes, that's correct.

So 9,999 = 9 * 11 * 101.

Therefore, 1331 is 11^3, and 9,999 is 9 * 11 * 101.

So the GCD of 1331 and 9,999 is 11. Because 11 is the common prime factor.

Therefore, we can divide both numerator and denominator by 11:

1331 ÷ 11 = 121

9,999 ÷ 11 = 909

So the simplified fraction is 121/909.

Now, check if 121 and 909 have any common factors. Let's factorize them.

121 is 11^2, as before.

909: Let's divide by 9 first. 9*101=909. Because 9*100=900, so 900+9=909. So 909=9*101.

Therefore, 909 = 3^2 * 101.

121 is 11^2, and 909 is 3^2 * 101. There are no common prime factors here. So the GCD is 1, meaning the fraction 121/909 is in its simplest form.

Therefore, a = 121 and b = 909. Adding them together: 121 + 909 = 1030.

Wait, let me double-check all steps to make sure I didn't make a mistake.

First step: Let x = 0.13311331...

Multiply by 10,000: 10,000x = 1331.1331...

Subtract x: 9,999x = 1331 => x = 1331/9999. That's correct.

Factorizing 1331: 11^3. Correct.

Factorizing 9999: 9*1111. 1111 is 101*11. So 9*11*101. Correct.

GCD of 11^3 and 9*11*101 is 11. Divide numerator and denominator by 11: 1331/11=121; 9999/11=909. Correct.

121 is 11^2, 909 is 9*101. No common factors. So a=121, b=909. Sum is 1030. That seems right.

But let me confirm with another method. Maybe using a different approach to converting the repeating decimal.

Alternatively, since the repeating part is four digits, maybe I can write the decimal as the sum of its digits multiplied by appropriate powers of 10.

But actually, the standard method I used earlier should be correct. Let me verify the decimal expansion of 121/909.

Calculating 121 divided by 909:

121 ÷ 909. Let's do this division.

909 goes into 121 zero times. Add a decimal point. 909 into 1210. 909*1=909. 1210 - 909 = 301. Bring down a 0: 3010. 909*3=2727. 3010-2727=283. Bring down a 0: 2830. 909*3=2727. 2830-2727=103. Bring down a 0: 1030. 909*1=909. 1030-909=121. Bring down a 0: 1210. This is the same as the first step after the decimal. So the decimal repeats: 0.13311331...

Yes, that matches the original decimal. So the fraction is correct.

Therefore, the answer is 121 + 909 = 1030.
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