(v) What is the least number of saplings that can be arranged in rows of 12, 15 or 40 in each row?
Answers
Answer:
We have to find the LCM of the given rows 12, 15 and 40. (Taking Highest Powers). Therefore, least number of saplings that can be arranged in given rows are 120.
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Answer:
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A number of saplings is such that when 3, 5, 6, 8, 10 or 15 saplings are planted in each row, every time two saplings are left out. What is the minimum number of saplings?
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3 Answers
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Bernard Ishiekwen, MBBS Medicine and Surgery, University of Calabar, Calabar, Nigeria (2023)
Answered 2 years ago · Author has 217 answers and 1.6M answer views
This question is an application Lowest Common Factor (LCM).
In short, the lowest common factor is the smallest number, a group of other numbers can divide completely.
In order to solve this question, we have to find the lowest common factor of 3, 5, 6, 8, 10, and 15.
In case you are not familiar with finding lowest common factors, what we do is that we keep dividing the set of numbers with prime factors starting with 2.
If a prime factor cannot divide a number, we leave it alone. At any point none of the numbers can be divided by the prime factor, we move to the next usable prime factor.
We stop when all the numbers have been divided down to 1.
In the end, we multiply all the prime factors we used in the division together to find the Lowest Common Factor.
First we use 2 to divide the numbers,
2 - (3/2) (5/2) (6/2) (8/2) (10/2) (15/2)
ans - 3 5 3 4 5 15
we are left with 3, 5, 3, 4, 5, 15. Next we use 2 to divide the numbers again,
2 - (3/2) (5/2) (3/2) (4/2) (5/2) (15/2)
ans - 3 5 3 2 5 15
we are left with 3, 5, 3, 2, 5, 15. There is still an even number so we use 2 again,
2 - (3/2) (5/2) (3/2) (2/2) (5/2) (15/2)
ans - 3 5 3 1 5 15
At the end of the third division, we are left with 3, 5. 3, 1, 5, 15. There is no even number left so we move to the next prime factor, 3.
3 - (3/3) (5/3) (3/3) (1/3) (5/3) (15/3)
ans - 1 5 1 1 5 5
After 3 divides, we are left with 1, 5, 1, 1, 5, 5. The next usable prime factor is 5.
5 - (1/5) (5/5) (1/5) (1/5) (5/5) (5/5)
ans - 1 1 1 1 1 1
Here is where we stop. So the LCM of 3, 5, 6, 8, 10 and 15
= 2 * 2 * 2 * 3 * 5
= 120
Now the LCM of 3, 5, 6, 8, 10 and 15 is 120. Meaning that 120 is the least number that can be divided by all those numbers. If we know this, then we know that if two saplings remain every time, the least number of sapling
= 120 + 2 = 122 saplings
Step-by-step explanation:
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