find hcf by alroathm of 418 and 33
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since 418>33 applying Euclid division algorithm
418= 33×12+22
since remainder not zero
applying EDL b/w 33 and 22
33= 22×1 +11
since remainder is not zero
Applying EDL b/w 22 &11
22= 11×2+0
so remainder is zero EDL stops
HCF (418,33)= 11 Ans
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418= 33×12+22
since remainder not zero
applying EDL b/w 33 and 22
33= 22×1 +11
since remainder is not zero
Applying EDL b/w 22 &11
22= 11×2+0
so remainder is zero EDL stops
HCF (418,33)= 11 Ans
Hope it helps You
Mark it brainliest
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