एकाधिकेन पूर्वेण

Ekadhikena Pūrveṇa

"By one more than the previous one"

LearnStep-by-step animation PracticeInteractive problems ProofWhy it works

What this sutra solves

Reach for this whenever you need to square a number ending in 5 — or write a fraction as a repeating decimal.

You are tiling a square room that measures 35 ft on every side and want the area at a glance.

becomes

35 \times 35

⚡

3 × 4 = 12, tack on 25 → 1225 sq ft, done in your head.

A stadium section has 65 rows of 65 seats and you need the capacity fast.

becomes

65^2

⚡

6 × 7 = 42, append 25 → 4225 seats.

Live Demo— animated step-by-step

math problem

²=?

direct way

35 × 35

long multiplication, carries, more steps

spot pattern

vedic cue

ends in 5

use 3 × 4, then append 25

Start with the square

The math problem is 35². A direct approach asks for 35 × 35; the Vedic shortcut notices that 35 ends in 5.

1 / 6

⚡ Speed Advantage

Vedic

2 steps

Traditional

7 steps

4× faster with Vedic Mathematics

Best for

• Squaring numbers ending in 5
• Computing recurring decimals

Use when

• Number ends in 5
• Computing 1/(10n+9) recurring decimals

Avoid when

• Numbers not ending in 5 (use Nikhilam or Urdhva instead)

Intuition

When the last digit is 5, the prefix does all the work — multiply it by itself plus one.

Story Mode

The Pattern in Every Square

A Vedic teacher once posed a riddle: 'Find a number whose square always ends in 25.' The answer: every number ending in 5. But why? Algebra reveals: ( $10a+5$ )² $= 10$ 0a( $a+1$ ) + 25. The '25' is locked in — it is the square of 5 itself. And a( $a+1$ )? That's the prefix multiplied by one-more-than-itself. The ancient pattern encodes a universal algebraic identity.

Vedic vs conventional

Conventional

$85^2 = 85\times 85$ via long multiplication (6+ steps, carrying).

Vedic

$8\times 9 = 72$ , append $25 \to 7225$ (1 mental step).

1 step vs 6+ steps for squaring numbers ending in 5.

Applications

Squaring numbers ending in 5

Any number ending in 5 can be squared in one mental step.

25^2

35^2

65^2

85^2

125^2

995^2

Recurring decimals of fractions

Fractions like 1/19, 1/29, 1/49 have long repeating cycles — generate them digit by digit using the multiplier.

1/19

1/29

Common Mistakes to Avoid

Forgetting to always append 25 at the end

Wrong approach

35^2

→

3\times 4=12

→writing 12 (wrong)

Correct approach

✓

35^2

→

3\times 4=12

→append 25

→1225

Why this happens

💡 Students remember the multiplication but forget the fixed ending.

Using wrong prefix for 3-digit numbers

Wrong approach

125^2

→prefix is 12, one-more is 13

→

12\times 13=156

→15625

Correct approach

✓

1Prefix is everything before the 5. For 125, prefix=12.

12\times 13=156

, append 25

→15625.

Why this happens

💡 Students use only the tens digit as prefix instead of all digits before 5.

Why It Works

Let n be a number ending in 5 with prefix a:

n = 10a + 5, \quad a \in \mathbb{Z}^+

Square n:

n^2 = (10a + 5)^2 = 100a^2 + 100a + 25

Factor out 100:

= 100 \cdot a(a+1) + 25

Identify the two parts:

n^2 = \underbrace{a \times (a+1)}_{\text{prefix} \;\times\; \text{(prefix + 1)}} \times 100 \;+\; \underbrace{25}_{\text{always fixed}}

∴ The last two digits are always 25. The left part is always a × (a+1) — the prefix multiplied by one more than itself.

For numbers ending in 5: $n = 1$ $0a + 5$ . Then $n^2 = (10a+5)$ ² $= 10$ $0a^2+10$ $0a+25 = 10$ 0a( $a+1$ )+25. The last two digits are always 25; the left part is $a\times (a+1)$ — 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.

Explore Next

Related Sutra

Nikhilaṃ Navataścaramaṃ Daśataḥ

All from 9 and the last from 10

Related Sutra

Yāvadūnaṃ

Whatever the deficiency

Related Sub-Sutra

Ānurūpyeṇa

Proportionately