using Euclid's division algorithm, find HCF of 441, 567, 693
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The Euclidean Algorithm for finding HCF (A,B) is as follows:
The Euclidean Algorithm for finding HCF (A,B) is as follows:If A=0 then HCF (A,B)=B, since the HCF (0,B)=B, and we can stop.
The Euclidean Algorithm for finding HCF (A,B) is as follows:If A=0 then HCF (A,B)=B, since the HCF (0,B)=B, and we can stop. If B=0 then HCF (A,B)=A, since the HCF (A,0)=A, and we can stop.
The Euclidean Algorithm for finding HCF (A,B) is as follows:If A=0 then HCF (A,B)=B, since the HCF (0,B)=B, and we can stop. If B=0 then HCF (A,B)=A, since the HCF (A,0)=A, and we can stop. Write A in quotient remainder form (A=BQ+R)
The Euclidean Algorithm for finding HCF (A,B) is as follows:If A=0 then HCF (A,B)=B, since the HCF (0,B)=B, and we can stop. If B=0 then HCF (A,B)=A, since the HCF (A,0)=A, and we can stop. Write A in quotient remainder form (A=BQ+R)Find HCF (B,R) using the Euclidean Algorithm since
The Euclidean Algorithm for finding HCF (A,B) is as follows:If A=0 then HCF (A,B)=B, since the HCF (0,B)=B, and we can stop. If B=0 then HCF (A,B)=A, since the HCF (A,0)=A, and we can stop. Write A in quotient remainder form (A=BQ+R)Find HCF (B,R) using the Euclidean Algorithm since HCF (A,B)=HCF(B,R)
The Euclidean Algorithm for finding HCF (A,B) is as follows:If A=0 then HCF (A,B)=B, since the HCF (0,B)=B, and we can stop. If B=0 then HCF (A,B)=A, since the HCF (A,0)=A, and we can stop. Write A in quotient remainder form (A=BQ+R)Find HCF (B,R) using the Euclidean Algorithm since HCF (A,B)=HCF(B,R)Here, HCF of 441 and 567 can be found as follows:-
The Euclidean Algorithm for finding HCF (A,B) is as follows:If A=0 then HCF (A,B)=B, since the HCF (0,B)=B, and we can stop. If B=0 then HCF (A,B)=A, since the HCF (A,0)=A, and we can stop. Write A in quotient remainder form (A=BQ+R)Find HCF (B,R) using the Euclidean Algorithm since HCF (A,B)=HCF(B,R)Here, HCF of 441 and 567 can be found as follows:-567=441×1+126
⇒ 441=126×3+63
⇒ 441=126×3+63⇒ 126=63×2+0
⇒ 441=126×3+63⇒ 126=63×2+0Since remainder is 0, therefore,
⇒ 441=126×3+63⇒ 126=63×2+0Since remainder is 0, therefore, H.C.F of (441,567) is =63
⇒ 441=126×3+63⇒ 126=63×2+0Since remainder is 0, therefore, H.C.F of (441,567) is =63Now H.C.F of 63 and 693 is
⇒ 441=126×3+63⇒ 126=63×2+0Since remainder is 0, therefore, H.C.F of (441,567) is =63Now H.C.F of 63 and 693 is693=63×11+0
⇒ 441=126×3+63⇒ 126=63×2+0Since remainder is 0, therefore, H.C.F of (441,567) is =63Now H.C.F of 63 and 693 is693=63×11+0Therefore, H.C.F of (63,693)=63
⇒ 441=126×3+63⇒ 126=63×2+0Since remainder is 0, therefore, H.C.F of (441,567) is =63Now H.C.F of 63 and 693 is693=63×11+0Therefore, H.C.F of (63,693)=63Thus, H.C.F of (441,567,693)=63.
Answer: 63
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