What Is Ekādhikena Pūrveṇa? — The Sutra Behind Two Mental Shortcuts

Ekādhikena Pūrveṇa — "by one more than the previous one" — is the sutra behind two of Vedic Math's best mental shortcuts: squaring numbers ending in 5, and multiplying by 11. The same idea applied to two patterns.

TL;DR: Ekādhikena Pūrveṇa — "by one more than the previous one" — is the Vedic Math sutra that drives two of the most-used mental shortcuts: squaring any number ending in 5, and multiplying by numbers whose complement to 10 is 1 (such as ×11, ×101, ×1001 via related extensions). For 85²: take the "previous" digit (8), multiply by one more than itself (9) to get 72, append 25 → 7225. For 65²: 6 × 7 = 42, append 25 → 4225. The underlying identity is (10a + 5)² = 100a(a + 1) + 25.

Ekādhikena Pūrveṇa — the phrase translates to "by one more than the previous one" — is the single Vedic sutra most responsible for the "did you just do that in your head?" moment that converts sceptics. It is the technique behind two of the clearest wins in mental arithmetic: squaring any number ending in 5, and the base-side rule for multiplying by 11.

The pedagogy point most explanations miss is that these two tricks are not separate tricks. They are the same sutra applied to two different number patterns. Once a child sees the connection, both techniques install for the price of one.

The sentence I hear every cohort

"Wait — is that it?"

On Day 4 of the Charter Cohort, after I walk a nine-year-old through 65² ("6 times 7 is 42, write 25, done — 4225"), the sentence I hear roughly four times out of five is some version of "wait, is that it?" Not disbelief exactly — more a quiet suspicion that something has to be wrong, because they just squared a two-digit number before their parent has finished reading the problem aloud. I used to try to explain the identity underneath immediately. I stopped. The algebra lands better three weeks later, once their body knows the trick works. The moment of surprised silence is the pedagogical event; the "why" chases it, not the other way around.

The two applications in one sentence each

Squaring numbers ending in 5: take the digit (or digits) to the left of the 5 — call it the "previous". Multiply it by one more than itself. Append 25. That is the complete method.

Multiplying by 11 (base-side extension): for a two-digit number N × 11, the leading digit of the answer is the leading digit of N (the "previous"); the units digit is the units of N. The middle is the sum, which equals the "previous" value plus the "one more". For 47 × 11: write 4 (previous), 4 + 7 = 11 (sum of previous and one-more), carry 1 → 517.

The connection: both techniques rely on the single idea that "if you know the previous, the one-more is one add away" — which is why "by one more than the previous" is the sutra's full name.

Squaring 5-endings — the full worked example

Compute 85² mentally.

Identify the "previous". The digit left of 5 is 8. (Previous = 8.)
Multiply by one more than itself. 8 × 9 = 72. (One-more of 8 = 9.)
Append 25. Write 72 then 25: 7225.

Three steps, about 4 seconds. The standard column method (85 × 85 = 85 × 80 + 85 × 5) takes 25 seconds and has three places to drop a carry. The gain is 6x on this problem — modest by Vedic standards, but it compounds because squaring 5-endings shows up everywhere.

Why it works — the algebra

If the number is (10a + 5), its square is:

(10a + 5)² = 100a² + 100a + 25 = 100a(a + 1) + 25

The "100a(a + 1)" is the write-the-product-of-previous-and-one-more step; the "+ 25" is the append 25 step. The identity is one of the standard algebraic manipulations students meet in grade 8 or 9. Vedic Math exposes the pattern as a one-step mental procedure rather than a FOIL expansion followed by rearrangement.

The three-digit case — 115² and beyond

The method scales. For 115²: the "previous" is 11. One-more of 11 is 12. 11 × 12 = 132. Append 25: 13225. Six seconds versus roughly 45 for the standard method. For 125²: 12 × 13 = 156, append 25 → 15625. For 205²: 20 × 21 = 420, append 25 → 42025.

The ceiling of the method is "any number ending in 5 where you can mentally compute 'previous × one-more'". For a four-digit input (e.g., 1005²), that mental multiplication (100 × 101 = 10100) is already two-digit × three-digit territory, where Urdhva-Tiryagbhyam is the natural tool. The Ekādhika framework still structures the answer; a second sutra handles the inner multiplication.

Three squares to try without a calculator

75² (7 × 8 = 56 → 5625). 95² (9 × 10 = 90 → 9025). 135² (13 × 14 = 182 → 18225). If all three took under 15 seconds total, Ekādhika is installed on 5-endings. The second trick (×11 base-side rule) uses the same "previous × one-more" instinct — see below.

Open the free squaring lesson →

The ×11 application — same sutra, different pattern

For a two-digit number N × 11, the pattern:

Leading digit of answer: leading digit of N (plus any carry — see below).
Middle digit: sum of the two digits of N.
Units digit: units digit of N.

For 47 × 11: write 4, then 4 + 7 = 11 so write 1 carry 1, then 7 → 4+1 | 1 | 7 → 517. The "by one more" connection: the final leading digit is one more than the original leading digit whenever the middle sum overflows. In the clean case (no overflow, e.g., 34 × 11 → 374), the leading digit stays as the "previous"; in the overflow case (47 × 11 → 517), it becomes "one more than the previous."

For a longer treatment of ×11 and its extensions (×111, ×1001, 3-digit × 11), see the complete ×11 field guide.

Where Ekādhika shows up in exam math

Squaring numbers ending in 5 is one of the most frequent sub-tasks on standardised tests, usually hidden inside a larger problem.

SAT: quadratic completion where the midpoint is a 5-ending number. Difference-of-squares problems like 75² − 25² where both terms are 5-endings reduce to Ekādhika on each term in about 8 seconds total.
ACT: geometry area with 25, 35, 45, 55, 65-side squares. Any statistics problem where variance involves a 5-ending mean.
JEE / CAT / SAT II / A-level: permutations, binomial coefficients, probability where sample sizes are 5-ending (particularly in combinatorics where C(10,5), C(15,5) involve numerous ×5 and 5² steps).
UK 11+: number-pattern recognition questions where identifying the sequence 25, 225, 625 as 5², 15², 25² is the whole answer.

The honest estimate: Ekādhika saves 15–30 seconds per fire, and a 5-ending-square fires roughly once every six questions on a mixed arithmetic/geometry section. That compounds to around 90 seconds per 60-question test — enough to matter.

Where Ekādhika does not apply

Two cases:

Squaring numbers not ending in 5. 83² is not an Ekādhika problem. Use Yavadunam (near-base squaring) if the number is close to a round base (83 is close to 100 — Yavadunam gives 6889 in 6 seconds). Otherwise use Urdhva-Tiryagbhyam.
Multiplying by 11 for numbers with three or more digits. The two-digit × 11 rule above works clean; three-digit × 11 (e.g., 456 × 11) requires an extended version with carries at both middle positions. See the ×11 field guide for the extension.

The install — two days for both applications

Ekādhika is the quickest Vedic sutra to install because the pattern is visually obvious — the "5 at the end" cue triggers the method unambiguously. Two days of 15 minutes typically does it for a child aged 10 or older.

Day 1: squaring 5-endings. 15 problems. Start with single-digit prefixes (15², 25², 35²…), move to two-digit prefixes (105², 115², 125²).
Day 2: ×11 rule for two-digit numbers. 15 problems. Include the overflow case (middle sum ≥ 10) from the 6th problem onward.

By the end of day two, both techniques fire automatically on recognition. The free 5-Day Mental Math Quick Start structures this as Day 1 (×11) and Day 4 (squaring 5-endings); the "by one more than the previous" connection is called out explicitly between them.

The Quick Start below is the install plan. Takes 20 minutes a day. Covers Ekādhika plus the three other most-used sutras over five days.

Frequently asked

What does Ekādhikena Pūrveṇa mean?

"Ekādhikena" = by one more than. "Pūrveṇa" = the previous one. The compound Sanskrit phrase describes the operational rule: for whichever quantity is "the previous one" in the setup, multiply or operate using one-more-than-that. Which quantity counts as "previous" depends on the application — for squaring 5-endings, the "previous" is the digit left of the 5; for the ×11 overflow case, the "previous" is the leading digit of the input.

What is the formula for squaring numbers ending in 5?

(10a + 5)² = 100a(a + 1) + 25. For 85², a = 8, so 100 × 8 × 9 + 25 = 7225. The method writes this as "a × (a+1) then append 25" — same identity, re-ordered for mental execution. It is on the standard algebra syllabus for grade 8-9.

Does Ekādhika work for numbers ending in digits other than 5?

No for the exact "append 25" form. The pattern (10a + 5)² is specifically engineered around the 5-and-5 identity giving a clean 25 append. For other endings, use Yavadunam (near-base squaring) if the number is close to a round base, or Urdhva-Tiryagbhyam otherwise.

Is the ×11 trick the same as the squaring trick?

Different applications, same sutra. Both rely on the "by one more than the previous" structure. For squaring 5-endings, "previous × one-more" IS the full computation. For ×11, the "one-more" shows up when the middle sum overflows and the leading digit of the answer becomes one more than the leading digit of the input.

How much time does Ekādhika save on an exam?

Honest estimate: 15–30 seconds per fire. A 5-ending-square fires roughly once every six questions on a standardised math test, so about 90 seconds per 60-question section. ×11 adds another 30–45 seconds across 2–3 fires per section. Combined, Ekādhika alone contributes about 2 minutes of recovered time — meaningful but not the biggest Vedic lever (Nikhilam saves more where it applies).

Install Ekādhika in the free 5-Day Quick Start → — Day 1 is ×11. Day 4 is 5-ending squares. Same sutra. Free.
https://www.thevedicmind.com/quick-start