Petals of a Flower

After Betty and Daisy removed 3 petals, the number of petals left is a multiple of 2.

Hence, the number of petals this beautiful flower have is odd.

So when Daisy removes 4 more petals from the flower, 10 petals would have been removed in total.

Therefore, the number of petals left will still be odd. (Odd - Even)

Hence, the number of petals left is not a multiple of 4.

Once Daisy removed 1 petal, it didn't give us any crucial information. However, she did this before removing two more petals, which was stated to be a multiple of 2. So after removing (1+2)=3 petals, the number of petals remaining was even.

This means the number of original petals was an even number + 3 = an odd number.

So after she removes (1+2+3+4)=10 petals, the number of petals remaining must be the original odd number - 10 = another odd number. Obviously, there are no odd numbers that are divisible by 4.

Therefore the number of petals remaining cannot be a multiple of 4.

Just for fun let's look at some examples that satisfy the first three criteria. If you plug in 15, you will find that it is an example. There are actually infinitely many examples. Let's list the next few examples: 21, 27, 33, 39, etc. We can prove that all numbers in the form of 6k+3, where k is an integer, is an example, provided that k>1. (If you want to prove that it is an example, you can try checking from the steps above for the flower petal). These numbers are actually the only examples for this question (By "only", I mean the infinitely many examples).

So the number of petals remaining cannot be a multiple of 4, and from the information already provided, the number of petals is 6*(an integer>1) +3.

Let's * suppose Daisy can remove the final four petals leaving a multiple of four * and then go on to get a contradiction.

When three petals have been removed we have a multiple of 2 left, so we can say

$n-3=2p$

When a total of ten petals (1+2+3+4) have been removed we are left with a multiple of 4, so we can say

$n-10=4q$

Now solve each of these equations for n and equate the answers to get

$2p+3=4q+10$

But this is impossible! Because the LHS is odd and the RHS is even.

Concluding our proof by contradiction we must deny our supposition and say

$\boxed{\text{ Daisy cannot remove the final four petals leaving a multiple of four }}$

If the number of petals is $x$ , then $x-1-2=x-3$ must be even. So $x-1-2-3-4=x-10$ must be odd, and odd number may not be divisible by $4$ .

The value in step 3 must be a multiple of 12 as it's divisible by 3 and 4. The difference between steps 2 and 3 is odd (3), which makes step 2 odd, but we know it's even. This is a contradiction, so the premise is false.

n-1 = 1 x
n-3 = 2
y => n is odd.
n-6 = 3*z => n is a multiple of 3
n-10 = 2y-7 ,which is odd and can't be a multiple of 4

take the number of petals that this flower as n. n can be either even or odd. when 1 is taken from n, n is divisible by 1, which is obvious. when 2 further is taken by n, it is divisible by 2, which means the result must be even. when we add an odd number to an even number, the result is always odd.

when 3 is taken from n, it is a multiple of 2, therefore the n-3 is even. But since it is even, and you add 3, an odd number to it, you get that n is odd. when 4 petals are plucked out of it further, the total number taken away from n is 10. An odd number minus an even number is odd, therefore since n-10 is odd, the number of petals remaining can't be divisible by 4.

We may assume the number of petal in order to create a simple solution! We will start from the final step!

In the final step the remaining petal is the multiple of 3! we assume its 3! Thus, before removing the last one there should be 4! and it is a multiple of 2! And the first one is a multiple of 1, it have to be correct!

Now back to the rest of this solution! If we remove another petal the ramaining petal is 2! And it couldn't be a multiple of 4!

So the answer is NO

The number of petals in tha flower can be the L.C.M of 1,2 &3or it's multiples. For example,- the number can be 6 which is not a multiple of 4 whereas it can also be 12 then it is a multiple of 4.

If we consider the number 15...we can easily see the solution before us

When Betty removes l Petal of the flower ,the flower lost a petal and is a multiple of 1 which isn’t enough information because we need to find the amount of petal is odd or even. When daisy removed 2 more petal, together the petal lost 3 petal and now the flower has a multiple of 2.

2times2 =4 2times3 =6 2times4 =8 2times5 =10 2times6 =12

Any of this numbers added to 3 will equal an odd number. When Betty again removed 3 more petals altogether 6 petals are removed and it’s still multiple of 3.

3times2 =6 3times3 =9 3times4 =12

So when we add 6 with any multiple of 3 it will be odd. When daisy removes 4 more petals it can’t be multiple of 4 because no multiple of 4 is an odd number

And the answer might be 15

Suppose that the remaining petals are a multiple of 4, i.e. 4k. Then one move before it was 4k+3, a muliple of 3. One move before that it was 4k+5, a multiple of 2 which is impossible, because it is an odd number. So our starting assumption is false.

Let the number of petals be 'x'. If I remove 3 petals (1 and 2 more) I am left with a multiple of 2. That is (x - 3) is a multiple of 2. Which means x is a odd number. Now after removing 7 more petals ( 3 + 4 ) I will be left with (x - 10). Since x is a odd number, (x-10) will also be an odd number. Which means it can't be a multiple of 4.

I just simplified it, suppose the flower has 100 petals.

1 removed ➡️ left 99 = multiple of 1

2 more removed➡️ left 97 🚫 multiple of 2 (🚫 Means not equal)

We can prove this by the method of contradiction.

Suppose after removing $1, 2, 3$ and $4$ number of petals, we get $4n$ number of petals. Where $n$ is a positive integer.

Then according to the question, $4n + 4$ number of petals should be a multiple of 3, i.e. $4(n+1)$ is divisible by 3. Since 4 is not divisible by 3, $(n+1)$ should be divisible by 3 which is possible for $n= 2,5,8. \dots$

Again after adding 3 more petals we get $4(n+1) +3$ number of petals, which should be divisible by 2. Since 4 is a multiple of 2, $4(n+1)$ is also a multiple of 2. So for $4(n+1) +3$ number of petals to be divisible by 2, 3 should be divisible by 2, which is not possible. Hence after removing $1, 2, 3$ and $4$ number of petals it is impossible to have $4n$ number of petals.

8 is the only number which is a multiple of 4 and if you add 4 it will be a multiple of 3. So 8+4=12 which is a multiple of 3 and 12+3=15 which is not a multiple of 2. So there will be no such number for the above question.

Ez ;) When 2 petals are removed the number of petals remaining is a multiple of 2, i.e., it's even. Before this 1 petal was removed. Which means to begin with, we had an odd number of petals. Now, after removing the next 3 petals, a total of 3+2+1=6 petals have been removed. The number of petals left behind has got to be odd. (Odd-Even=Odd) Now, next after removing 4 petals, the number of petals remaining will still be odd. (Odd-Even=Odd, remember?). But 4 definitely cannot divide an odd number. So, no, that's not possible.

Winston, nice footnote, dude.