sahiti220620
Bunuel can you please show that conversion to 1/99 fraction here and solve
sahiti220620 See if the below explanation helps:
Understanding the 1/99 PatternFirst, recall that \(\frac{1}{99} = 0.\overline{01}\) (01 repeats)
More generally: \(\frac{N}{99} = 0.\overline{N}\) where N is a 2-digit number that repeats.
For example: \(\frac{36}{99} = 0.\overline{36} = 0.363636...\)
Converting 13/55 Using This TechniqueStep 1: Find the decimal representation
\(\frac{13}{55} = 0.2\overline{36}\)
Step 2: Separate the non-repeating and repeating parts
\(0.2\overline{36} = 0.2 + 0.0\overline{36}\)
Step 3: Express the repeating part using 1/99
\(0.0\overline{36} = \frac{0.\overline{36}}{10} = \frac{36/99}{10} = \frac{36}{990} = \frac{4}{110}\)
Step 4: Verify
\(\frac{13}{55} = \frac{2}{10} + \frac{4}{110} = \frac{22}{110} + \frac{4}{110} = \frac{26}{110} = \frac{13}{55}\) ✓
Finding the 57th DigitNow we know: \(\frac{13}{55} = 0.2\overline{36}\)
- Position 1: \(2\) (non-repeating)
- Positions 2 onwards: \(36\) repeats
For position 57:
- We need the \((57-1) = 56^{th}\) digit of the repeating pattern "\(36\)"
- \(56 \div 2 = 28\) with remainder \(0\)
- This means we complete exactly 28 cycles of "\(36\)"
- The last digit of each cycle is \(6\)
Answer: E (6)Pattern Recognition TechniqueWhen you see a fraction that produces a repeating decimal, immediately think:
- 1-digit repeat → denominator with factor 9
- 2-digit repeat → denominator with factor 99
- 3-digit repeat → denominator with factor 999
This \(\frac{N}{99}\) technique works whenever you can isolate the repeating part!