using euclid's division algorithm find whether the HCF of 2160 and 3520
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The HCF OF 2160 & 3520 IS 80.
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REGARD
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Euclid's division algorithm:
According to this the HCF of any two positive integers a and b, with a>b
a= bq+r, 0 ≤r
_____________________________
a= 3520 , b= 2160
On applying euclid's division Lemma for 2160 & 3520
3520 = (2160×1)+1360
Here, remainder= 1360≠0
So take new Dividend as 2160 & divisior as 1360
2160 = (1360×1)+800
Here, remainder= 800≠0
So take new Dividend as 1360 & divisior as 800
1360= (800×1)+560
Here, remainder= 560≠0
So take new Dividend as 800 & divisior as 560
800 = (560×1)+240
Here, remainder= 240≠0
So take new Dividend as 560 & divisior as 240
560 = (240×2)+80
Here, remainder= 80≠0
So take new Dividend as 240 & divisior as 80
240 = (80×3)+0
Here, the remainder is 0 & last divisor is 80.
Hence , HCF of 2160 & 3520 is 80
==================================================================
Hope this will help you....
According to this the HCF of any two positive integers a and b, with a>b
a= bq+r, 0 ≤r
_____________________________
a= 3520 , b= 2160
On applying euclid's division Lemma for 2160 & 3520
3520 = (2160×1)+1360
Here, remainder= 1360≠0
So take new Dividend as 2160 & divisior as 1360
2160 = (1360×1)+800
Here, remainder= 800≠0
So take new Dividend as 1360 & divisior as 800
1360= (800×1)+560
Here, remainder= 560≠0
So take new Dividend as 800 & divisior as 560
800 = (560×1)+240
Here, remainder= 240≠0
So take new Dividend as 560 & divisior as 240
560 = (240×2)+80
Here, remainder= 80≠0
So take new Dividend as 240 & divisior as 80
240 = (80×3)+0
Here, the remainder is 0 & last divisor is 80.
Hence , HCF of 2160 & 3520 is 80
==================================================================
Hope this will help you....
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