Use Euclid's Division Algorithm to find the HCF of- a) 405 and 2520, b) 960 and 1575.
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Answered by
135
Hi ,
Euclid's division lemma:
Let a and b any two positive integers . Then there exist two
unique q and r such that
a = bq + r ,
0 less or equal to ' r ' less than b.
a ) 405 and 2520 , start with the larger integer that is , 2520
Apply the division lemma to 2520 and 405 ,
2520 = 405 × 6 + 90
Since the remainder 90 , we apply the division lemma to
405 and 90
405 = 90 × 4 + 45
90 = 45 × 2 + 0
The remainder has now become zero, so procedure stops.
Therefore ,
HCF( 405 , 2520 ) = 45.
b) To find HCF of 960 and 1575
1575 = 960 × 1 + 615
960 = 615 × 1 + 345
615 = 345 × 1 + 270
345 = 270 × 1 + 75
270 = 75 × 3 + 45
75 = 45 × 1 + 30
45 = 30 × 1 + 15
30 = 15 × 2 + 0
Now remainder is equal to zero.
Therefore ,
HCF ( 960 , 1575 ) = 15
I hope this will useful to you.
*****
Euclid's division lemma:
Let a and b any two positive integers . Then there exist two
unique q and r such that
a = bq + r ,
0 less or equal to ' r ' less than b.
a ) 405 and 2520 , start with the larger integer that is , 2520
Apply the division lemma to 2520 and 405 ,
2520 = 405 × 6 + 90
Since the remainder 90 , we apply the division lemma to
405 and 90
405 = 90 × 4 + 45
90 = 45 × 2 + 0
The remainder has now become zero, so procedure stops.
Therefore ,
HCF( 405 , 2520 ) = 45.
b) To find HCF of 960 and 1575
1575 = 960 × 1 + 615
960 = 615 × 1 + 345
615 = 345 × 1 + 270
345 = 270 × 1 + 75
270 = 75 × 3 + 45
75 = 45 × 1 + 30
45 = 30 × 1 + 15
30 = 15 × 2 + 0
Now remainder is equal to zero.
Therefore ,
HCF ( 960 , 1575 ) = 15
I hope this will useful to you.
*****
Answered by
73
Euclid division lemma:-
a=bq+r
0≤r
(a)405 and 2520
2520 = 405 × 6 + 90
405 = 90 × 4 + 45
90 = 45 × 2 + 0
As the remainder is 0,the HCF of 405 and 2520 is 45
HCF (405,2520) = 45
(b)960 and 1575
1575 = 960 × 1 + 615
960 = 615 × 1 + 345
615 = 345 × 1 + 270
345 = 270 × 1 + 75
270 = 75 × 3 + 45
75 = 45 × 1 + 30
45 = 30 × 1 + 15
30 = 15 × 2 + 0
As the remainder is 0,HCF of 960 and 1575 is 15.
HCF (960,1575) = 15
★hope it helps★
a=bq+r
0≤r
(a)405 and 2520
2520 = 405 × 6 + 90
405 = 90 × 4 + 45
90 = 45 × 2 + 0
As the remainder is 0,the HCF of 405 and 2520 is 45
HCF (405,2520) = 45
(b)960 and 1575
1575 = 960 × 1 + 615
960 = 615 × 1 + 345
615 = 345 × 1 + 270
345 = 270 × 1 + 75
270 = 75 × 3 + 45
75 = 45 × 1 + 30
45 = 30 × 1 + 15
30 = 15 × 2 + 0
As the remainder is 0,HCF of 960 and 1575 is 15.
HCF (960,1575) = 15
★hope it helps★
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