Lesson 14: Vedic Mathematics — Ancient Speed Techniques

Vedic Mathematics is the system of speed techniques popularised by Swami Bharati Krishna Tirtha's 1965 book of the same name. Historians debate how ancient the *sutras* are; nobody debates that the techniques work, because each one is a theorem of ordinary algebra wearing a shortcut's clothes. This lesson teaches the four most useful sutras for Classes IX–X — with the algebra that proves them, because a technique you can justify is a technique you can trust under exam pressure.

Sutra 1: Nikhilam — "All from 9, last from 10" (multiplication near a base)

For numbers near 100, write each number's difference from 100. To compute 98 × 97: differences are −2 and −3. Cross-subtract: 98 − 3 = 95 (same as 97 − 2). Multiply the differences: (−2)(−3) = 6, written as two digits: 06. Answer: 9506. Above the base it works identically: 103 × 105: differences +3, +5; cross-add 103 + 5 = 108; 3 × 5 = 15; answer 10815. Why it works: (100 − a)(100 − b) = 100(100 − a − b) + ab. The cross-subtraction builds the bracket; the product of differences fills the last two digits. Nikhilam also subtracts instantly: 1000 − 457 — take each digit from 9 and the last from 10: 5, 4, 3 → 543.

Sutra 2: Urdhva-Tiryagbhyam — "Vertically and crosswise" (general multiplication)

For 23 × 41, work in three columns: vertical left 2 × 4 = 8; crosswise 2 × 1 + 3 × 4 = 14 (write 4, carry 1); vertical right 3 × 1 = 3. Combine: 8 | 14 | 3 → 943 ✓ (23 × 41 = 943). The pattern generalises to any size — the columns are exactly the pairwise products that long multiplication computes, reordered so partial results arrive in a carry-friendly stream. It is also precisely how polynomial multiplication works, which is why it reappears in algebra: (2x + 3)(4x + 1) by the same columns gives 8x² + 14x + 3.

Sutra 3: Ekadhikena Purvena — "By one more than the previous" (squares ending in 5)

To square a number ending in 5: multiply the leading part by *one more than itself*, then append 25. 65²: 6 × 7 = 42, append 25 → 4225. 85²: 8 × 9 = 72 → 7225. Algebra: (10a + 5)² = 100a(a + 1) + 25 — the proof is one line, and knowing it stops the classic slip of squaring the whole "one more" wrongly.

Sutra 4: Duplex squares

For squaring two-digit numbers generally, use duplexes: for 31², compute D(3) = 9, D(31) = 2 × 3 × 1 = 6, D(1) = 1 → 961 ✓. Combined with Nikhilam for numbers near a base, most two- and three-digit squares become mental arithmetic.

A carry-heavy Nikhilam (the honest hard case). 88 × 82: differences from 100 are 12 and 18. Cross-subtract: 88 − 18 = 70. Multiply differences: 12 × 18 = 216 — but only two digits are allowed in the right block, so carry the 2: 70 + 2 | 16 → 7216 ✓ (88 × 82 = 7216). The carry rule is where Nikhilam users err; the defence is a magnitude check: 88 × 82 must be near 90 × 80 = 7200, and 7216 is, while the uncarried misread "70216" obviously is not.

Competency application: the shopkeeper's bill

A wholesaler sells notebooks at ₹98 each; a school orders 19. Mental route: 19 × 98 = 19 × (100 − 2) = 1900 − 38 = ₹1862 — this is the *Anurupyena* (proportionate) spirit: rewrite one factor near a round base. A second bill: 97 notebooks at ₹103: 97 × 103 = 97 × 100 + 97 × 3 = 9700 + 291 = ₹9991. In the case set, examinees must *show* the sutra working, not just the answer — write the differences and the cross step; method marks are the point.

Common mistakes

• Forgetting the right block must have exactly as many digits as the base has zeros. For base 100, two digits: 6 becomes 06, and 216 forces a carry. Most Nikhilam errors are digit-count errors, not concept errors.
• Cross-adding when both numbers are below the base (or vice versa). Below base: cross-*subtract*; above base: cross-*add*; mixed (one above, one below) needs the signed-difference version — know its limits.
• Applying Ekadhikena to numbers not ending in 5. The append-25 trick is exclusive to the digit 5.
• Treating sutras as magic in proofs. In a "show that" question, expand the algebra as above — the sutra alone earns no proof marks.

Exam Experience

Section A — Multiple choice (1 mark each)

Q1. Using Ekadhikena Purvena, 105² equals: (a) 11025 (b) 10225 (c) 10025 (d) 11225. Answer: (a) 11025. 10 × 11 = 110, append 25.

Q2. By Nikhilam (base 100), 99 × 97 equals: (a) 9603 (b) 9503 (c) 9693 (d) 9306. Answer: (a) 9603. Cross-subtract 99 − 3 = 96; multiply differences 1 × 3 = 03.

Section B — Short answer (3 marks) Q3. Compute 88 × 82 using Nikhilam, showing the cross-subtraction and the carry. Marking scheme: 1 mark for the differences 12 and 18 from base 100; 1 mark for cross-subtraction 88 − 18 = 70 and product 12 × 18 = 216; 1 mark for carrying 2 into the left block to obtain 7216.

Section C — Case-based long answer (5 marks) Q4. A school canteen buys supplies in bulk. (i) 98 plates at ₹97 each. (ii) 103 cups at ₹105 each. (iii) The manager claims the plate bill exceeds the cup bill — verify or refute using sutra methods, showing all steps. (iv) Prove algebraically why Nikhilam works for numbers below base 100. Marking scheme: (i) Differences 2, 3; cross 98 − 3 = 95; product 06 → ₹9506 (1 mark). (ii) Differences 3, 5; cross 103 + 5 = 108; product 15 → ₹10815 (1 mark). (iii) 9506 < 10815, so the claim is false — the cup bill is larger (1 mark). (iv) (100 − a)(100 − b) = 10000 − 100a − 100b + ab = 100(100 − a − b) + ab (1 mark for expansion, 1 mark for identifying the bracket as the cross-subtraction and ab as the right block).

Going deeper: more sutras in the toolkit

Antyayordashake — "If the sum of the last digits is 10". When two 2-digit numbers share the same tens digit and their units sum to 10, multiply like Ekadhikena: 43 × 47 (tens 4, units 3 + 7 = 10): left part 4 × 5 = 20, right part 3 × 7 = 21, answer 2021 ✓ (43 × 47 = 2021). The algebra: (10a + b)(10a + 10 − b) with the shared structure gives 100a(a + 1) + b(10 − b) — the same "one more than the previous" heart as the 5-square trick.

Squares near 50 (a Nikhilam cousin). For 47²: write 47 = 50 − 3. Then 47² = (25 − 3) × 100 + 3² = 2200 + 9 = 2209 ✓. The pattern: (50 − k)² = (25 − k)·100 + k², and (50 + k)² = (25 + k)·100 + k², so 53² = 2800 + 9 = 2809 ✓. Base 50 works because (50)² = 2500 splits neatly into hundreds.

Multiplying by 11. Write the outer digits, insert pairwise sums between: 35 × 11 → 3 | 3 + 5 | 5 = 385; 47 × 11 → 4 | 11 | 7 → carry: 517 ✓. It is the distributive law (×10 + ×1) made visible — and it is also the first two rows of Pascal's triangle doing push-ups.

Going deeper: checking with digital roots (casting out nines)

Replace a number by the repeated sum of its digits (its digital root, with 9 for multiples of 9). Multiplication preserves digital roots. Check 98 × 97 = 9506: roots are 98 → 8 and 97 → 7; 8 × 7 = 56 → 5 + 6 = 11 → 2. And 9506 → 9 + 5 + 0 + 6 = 20 → 2 ✓. A mismatch *proves* an error; a match gives strong (not certain) evidence of correctness — one transposition of digits shares a digital root. Use it as the smoke alarm after any sutra computation: fast, catches most slips, costs five seconds.

Final drills

Drill 1 — Nikhilam at base 1000. Compute 998 × 997. Differences from 1000: 2 and 3. Cross-subtract: 998 − 3 = 995. Product of differences: 2 × 3 = 6, written in a *three*-digit block: 006. Answer: 995006 ✓ (998 × 997 = 995,006). The block width equals the number of zeros in the base — the same rule as base 100, one digit wider. Also 999² = (1000 − 1)² = 1,000,000 − 2,000 + 1 = 998001, obtainable by the same "all from 9" thinking.

Drill 2 — proportionate multiplication (Anurupyena spirit). 204 × 98: rewrite 98 as 100 − 2: 204 × 100 − 204 × 2 = 20,400 − 408 = 19,992. And 385 ÷ 11 by reversing the ×11 pattern: 385 has outer digits 3 and 5 with middle 8 = 3 + 5 ✓, so the quotient is 35. Each sutra in this lesson is bidirectional — multiplication tricks read backwards become division and factorisation checks.