19. An army contingent of 1000 members is to march behind an army band of 56 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march ?
Answers
Answered by
206
Hi ,
It is related to Eculid's Division Lemma,
__________________________________
Given positive integers a and b , there exist
unique integers q and r satisfying
a = bq + r ,
Where 0 less than or equals to r < b.
______________________________
According to the problem ,
Members in army contigent = 1000
Members in army band = 56
To find maximum number of columns to march
behind army band in the same number of columns,
we have to find HCF of 1000 and 56
1000 = 56 × 17 + 48
56 = 48 × 1 + 8
48 = 8 × 6 + 0
The remainder has become zero , so our procedure
stops .
Since the divisor at this stage is 8 .
HCF ( 1000 , 56 ) = 8
Required columns = HCF ( 1000 , 56 ) = 8
I hope this helps you.
****
It is related to Eculid's Division Lemma,
__________________________________
Given positive integers a and b , there exist
unique integers q and r satisfying
a = bq + r ,
Where 0 less than or equals to r < b.
______________________________
According to the problem ,
Members in army contigent = 1000
Members in army band = 56
To find maximum number of columns to march
behind army band in the same number of columns,
we have to find HCF of 1000 and 56
1000 = 56 × 17 + 48
56 = 48 × 1 + 8
48 = 8 × 6 + 0
The remainder has become zero , so our procedure
stops .
Since the divisor at this stage is 8 .
HCF ( 1000 , 56 ) = 8
Required columns = HCF ( 1000 , 56 ) = 8
I hope this helps you.
****
Answered by
45
Answer:
Step-by-step explanation:
- Number of army contingents = 1000
- Number of people in the army band = 56
1000 = 2 x 2 x 2 x 5 x 5 x 5
56 = 2 x 2 x 2 x 7
Therefore: HCF (1000,56) = 8
Therefore the maximum number of columns in which they can march is 8.
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