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Euclid's division lemma : Let a and b are two positive integers. There exist unique integers q and r such that
a = bq + r, 0 r < b
Or We can write it as,
Dividend = Divisor × Quotient + Remainder
<u>Work</u><u> </u><u>out</u><u>:</u>
Given integers are 240 and 228. Clearly 240 > 228. Applying Euclid's division lemma to 240 and 228,
⇛ 240 = 228 × 1 + 12
Since, the remainder 12 ≠ 0. So, we apply the division dilemma to the division 228 and remainder 12,
⇛ 228 = 12 × 19 + 0
The remainder at this stage is 0. So, the divider at this stage or the remainder at the previous age i.e 12
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