A nursery distributes an equal number of flowers every day so that no flowers are left undistributed in the field.
On the first day they were divided into pairs.
On the second day they were divided into 4 equal groups.
On the third day they were divided into 5 groups.
It is known that the number of flowers is greater than 19 and less than 29.
Calculate the number of flowers that are distributed in the nursery over the three days.
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A nursery distributes an equal number of flowers every day so that no flowers are left undistributed in the field.
On the first day they were divided into pairs.
On the second day they were divided into 4 equal groups.
On the third day they were divided into 5 groups.
It is known that the number of flowers is greater than 19 and less than 29.
Calculate the number of flowers that are distributed in the nursery over the three days.
To solve the problem, we first determine the LCM of the numbers 2, 4, and 5:
The LCM is determined by taking the highest power of each prime number: and , so .
Next, we consider numbers between 19 and 29: 20, 21, 22, 23, 24, 25, 26, 27, 28, 29.
Among these, only 20 is divisible by 20.
Therefore, the number of flowers is .
So, the solution to the problem is: .
Will a number divisible by 6 necessarily be divisible by 3?
The LCM (Least Common Multiple) gives us the smallest number that divides evenly by 2, 4, and 5. Since flowers must be distributed with no leftovers each day, we need a multiple of all three numbers!
Look at prime factorizations: 2 = 2¹, 4 = 2², 5 = 5¹. Take the highest power of each prime. Since 4 already contains 2², don't count the 2 from '2' separately!
We'd choose the smallest one since LCM gives the least common multiple. But here, only 20 works because the next multiple would be 40, which exceeds our range.
Yes! That's a valid approach. Check if each number divides evenly by 2, 4, and 5. But finding the LCM first is more efficient and helps you understand the mathematical structure.
Since 24 ÷ 5 gives a decimal, there would be leftover flowers on day 3, violating the problem conditions.
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