HCF of 488 and252 using Euclid's Division Algorithm.
Answers
Answer:
⭐⭐Here ur answer....... ⭐⭐
Given integers,
595 and 252
Applying Euclid division algorithm to 595 and 252 we get,
595=252×2+91.........................................1
Now applying ,Euclid division algorithm to 252 and 91 we get,
252=91×2+70.....….………...........................2
Now applying Euclid division algorithm to 91 and 70,we get
91=70×1+21…...............................................3
Now applying Euclid division algorithm to 70 and 21 we get,
70=21×3+7...................................................4
Now applying Euclid division algorithm to 21 and 7 we get,
21=7×3+0......................................................5
The remainder at this stage is zero.
Hence HCF of 595 and 252 is 7.
eq1,eq2,eq3, and eq4 can be written as,
595-(252×2)=91..........................................6
252-(91×2)=70.............................................7
91-(70×1)=21...................................................8
70-(21×3)=7.....................................................9
Now eq9 we have,
(as we express HCF in the form of a equation that is the reason we start from eq9)
7=70-(21×3)
By putting eq8 in above equation we have,
7=70-(91-70×1)3
7=70-91×3+70×3
7=70×4-91×3
By putting eq7 in above equation we have,
7=(252-91×2)4-91×3
7=252×4-91×8-91×3
7=252×4-91×11
By putting eq6 in above equation we have,
7=252×4-(595-252×2)11
7=252×4-595×11+252×22
7=252×26-595×11
7=595(-11)+252(26)
7=595m+252n, where m=-11 and n=26.
Hence HCF of 595 and 252 is in the form of 595m+252n,where m=-11 and n=26.
✌✌it may help you✌✌
First Of All We will divide 488 nd 252
let a=488 nd b=252
a is greater than b.....
so let's Start dividing
-------------so the Answer will be 252------------------