एकाधिकेन पूर्वेण — Ekadhikena Purvena
By one more than the previous one. When the last digit is 5, the prefix does all the work — multiply it by itself plus one.
Why it works: For numbers ending in 5: n = 10a + 5. Then n² = (10a+5)² = 100a²+100a+25 = 100a(a+1)+25. The last two digits are always 25; the left part is a×(a+1) — prefix times one-more-than-prefix. For recurring decimals of 1/(10n+9): the multiplier is (n+1), and repeatedly multiplying from the right generates the full repeating cycle.
Speed advantage: 1 step vs 6+ steps for squaring numbers ending in 5.
Best for
- Squaring numbers ending in 5
- Computing recurring decimals
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Originally published at https://www.sutraflow.app/sutras/ekadhikena-purvena
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