Ekādhikena Pūrveṇa
By one more than the previous one
Squaring numbers ending in 5, and converting fractions like 1/19, 1/29.
Squaring any number that ends in 5; computing recurring decimals of fractions whose denominator ends in 9.
Take the digit(s) before the final 5, multiply by 'one more than itself', then append 25. For 1/19, 1/29 etc., use the digit one more than the digit before 9 as a multiplier in a left-to-right or right-to-left build of the recurring block.
For …5²: prefix × (prefix + 1), then append 25.
Worked examples
Try it yourself
Try to solve, then reveal the steps one at a time.
Modern applications
Where this ancient technique still lives in today's world.
Bit-shift squaring
Squaring numbers ending in 5 maps to a left-shift + add pattern; useful for hash-mix routines and constant-time arithmetic in inner loops.
Quick interest checks
Mental squaring of mid-sized factors for compound-interest sanity checks without a calculator.
Code in your favourite language
Square any number ending in 5 using ‘by one more than the previous’.
# square_ending_in_5(35) -> 1225
def square_ending_in_5(n: int) -> int:
prefix = n // 10
return prefix * (prefix + 1) * 100 + 25
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