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एकन्यूनेन पूर्वेण

Ekanyūnena Pūrveṇa

"By one less than the previous one"

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What this sutra solves

Multiply anything by a string of nines (9, 99, 999) instantly.

Seven items priced at ₹99 each — total at the till?

becomes7×997 \times 99
⚡

One less than 7 is 6, complement is 93 → 693.

A bulk order of 43 units at ₹99 apiece.

becomes43×9943 \times 99
⚡

42 | 57 → 4257.

Live Demo
7
×
99
1

Identify the multiplier type

99 is a string of 2 nines, so use the "one less than before" shortcut.

1 / 5

⚡ Speed Advantage

Vedic
1 steps
Traditional
9 steps

9× faster with Vedic Mathematics

Best for

  • • Multiplying by numbers that are all 9s

Use when

  • • Multiplying by 9, 99, 999, 9999, etc.

Avoid when

  • • Other multipliers

Intuition

Multiply by a string of 9s: subtract 1 from the multiplicand for the left part; find the complement of the multiplicand for the right part.

Story Mode

The 9s Shortcut

Multiplying by 99 is the same as multiplying by 100 and subtracting once. The Ekanyunena sutra packages this insight: the left part is 'one less than the multiplicand'; the right part is what completes it to the next round number. Together they are the product.

Vedic vs conventional

Conventional

423×999423\times 999 via long multiplication — 9 partial products.

Ekanyunena

left=422, right=577→422577577 \to 422577 (1 step).

1 mental step vs 9+ for multiplying by 9, 99, 999, 9999.

Applications

Multiplying by 9, 99, 999, ...

Any number multiplied by a string of 9s in one step.

35×9935 \times 9942×942 \times 9423×999423 \times 9992345×99992345 \times 9999

Common Mistakes to Avoid

✕

Right part must have same digit count as the multiplier

Wrong approach

35×9935\times 99: right = complement of 35=6535 = 65. Correct — both 2-digit. For 5×995\times 99: right = complement of 05=9505 = 95.

Correct approach

✓
Zero-pad the multiplicand to match the count of 9s in the multiplier before taking complement.

Why this happens

💡 Students take complement of the number as-is without padding.

Why It Works

A string of k nines is one less than a power of 10:

99…9⏟k digits=10k−1\underbrace{99\ldots9}_{k\text{ digits}}=10^k-1

Multiply by that form:

n(10k−1)=n⋅10k−nn(10^k-1)=n\cdot10^k-n

Rewrite as left part and complement:

=(n−1)⋅10k+(10k−n)=(n-1)\cdot10^k+(10^k-n)

So the left part is one less than the number, and the right part is its complement to the next power of 10.

n×(10k−1)=n×10k−n=(n−1)n \times (10^k - 1) = n \times 10^k - n = (n-1) followed by complement(n, k). Because 10k−n10^k - n is the complement of n with respect to 10k10^k: writing n in k digits and subtracting from 10k10^k gives the right portion.

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Related Sutra

Nikhilaṃ Navataścaramaṃ Daśataḥ

All from 9 and the last from 10