Apply “buy X get Y” discount on N products
$begingroup$
Consider a product is added to a cart with quantity of N and "Buy X and get y" offer is applied i.e. Buy two get one free.
If n = 3 then I could simply apply a discount rate of 33.33 on each product.
But my problem occurs when there are more than 3.
If n = 4 then number of free products would still be 1 and I would have to pay full price for 4th product, but if I apply the above method I would be applying discount on the 4th product as well which should not be the case.
This continues for n = 4,5,6,7,8,9,.....
I need a formula to correctly calculate the price with the scheme applied.
Thank you.
Edit: Edited the question as it was not clear. Sorry.
X would be the number of products that needs to be purchased for the offer to apply and N would be the number of products that are actually in the cart.
algebra-precalculus percentages
asked Dec 24 '18 at 6:41
sdb1995sdb1995
113
$endgroup$
add a comment |
$begingroup$
Consider a product is added to a cart with quantity of N and "Buy X and get y" offer is applied i.e. Buy two get one free.
If n = 3 then I could simply apply a discount rate of 33.33 on each product.
But my problem occurs when there are more than 3.
If n = 4 then number of free products would still be 1 and I would have to pay full price for 4th product, but if I apply the above method I would be applying discount on the 4th product as well which should not be the case.
This continues for n = 4,5,6,7,8,9,.....
I need a formula to correctly calculate the price with the scheme applied.
Thank you.
Edit: Edited the question as it was not clear. Sorry.
X would be the number of products that needs to be purchased for the offer to apply and N would be the number of products that are actually in the cart.
algebra-precalculus percentages
asked Dec 24 '18 at 6:41
sdb1995sdb1995
113
$endgroup$
add a comment |
$begingroup$
Consider a product is added to a cart with quantity of N and "Buy X and get y" offer is applied i.e. Buy two get one free.
If n = 3 then I could simply apply a discount rate of 33.33 on each product.
But my problem occurs when there are more than 3.
If n = 4 then number of free products would still be 1 and I would have to pay full price for 4th product, but if I apply the above method I would be applying discount on the 4th product as well which should not be the case.
This continues for n = 4,5,6,7,8,9,.....
I need a formula to correctly calculate the price with the scheme applied.
Thank you.
Edit: Edited the question as it was not clear. Sorry.
X would be the number of products that needs to be purchased for the offer to apply and N would be the number of products that are actually in the cart.
algebra-precalculus percentages
asked Dec 24 '18 at 6:41
sdb1995sdb1995
113
$endgroup$
Consider a product is added to a cart with quantity of N and "Buy X and get y" offer is applied i.e. Buy two get one free.
If n = 3 then I could simply apply a discount rate of 33.33 on each product.
But my problem occurs when there are more than 3.
If n = 4 then number of free products would still be 1 and I would have to pay full price for 4th product, but if I apply the above method I would be applying discount on the 4th product as well which should not be the case.
This continues for n = 4,5,6,7,8,9,.....
I need a formula to correctly calculate the price with the scheme applied.
Thank you.
Edit: Edited the question as it was not clear. Sorry.
X would be the number of products that needs to be purchased for the offer to apply and N would be the number of products that are actually in the cart.
algebra-precalculus percentages
algebra-precalculus percentages
asked Dec 24 '18 at 6:41
sdb1995sdb1995
113
asked Dec 24 '18 at 6:41
sdb1995sdb1995
113
edited Dec 24 '18 at 7:04
sdb1995
asked Dec 24 '18 at 6:41
sdb1995sdb1995
113
asked Dec 24 '18 at 6:41
sdb1995sdb1995
113
asked Dec 24 '18 at 6:41
sdb1995sdb1995
113
113
add a comment |
add a comment |
1 Answer
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$begingroup$
The simple answer is if one orders $n$ of the product you get $lfloor frac n3 rfloor$ free and pay for $n-lfloor frac n3 rfloor$
answered Dec 24 '18 at 6:46
Ross MillikanRoss Millikan
299k24200374
$endgroup$
$begingroup$
I don't think that will work. The discount must be on an integral number of widgets. There would only be one free widget for four purchased. Please see the example in the question, then see my answer.
$endgroup$
– Adam Hrankowski
Dec 24 '18 at 6:52
1
$begingroup$
That is exactly the point of the floor function. You round the fraction down to the next whole number. It is exactly what you asked for-the last of three is free.
$endgroup$
– Ross Millikan
Dec 24 '18 at 6:58
$begingroup$
Oh, you are right. I read your answer too quickly and didn't catch the notation. I'll delete my answer as it is therefore a duplicate.
$endgroup$
– Adam Hrankowski
Dec 24 '18 at 7:01
$begingroup$
Thanks a lot. It worked. I've edited my question to clarify X and N.
$endgroup$
– sdb1995
Dec 24 '18 at 7:04
$begingroup$
So the general formula would be if N >= X then we can pay for n - floor(n/x)
$endgroup$
– sdb1995
Dec 24 '18 at 7:10
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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votes
$begingroup$
The simple answer is if one orders $n$ of the product you get $lfloor frac n3 rfloor$ free and pay for $n-lfloor frac n3 rfloor$
answered Dec 24 '18 at 6:46
Ross MillikanRoss Millikan
299k24200374
$endgroup$
$begingroup$
I don't think that will work. The discount must be on an integral number of widgets. There would only be one free widget for four purchased. Please see the example in the question, then see my answer.
$endgroup$
– Adam Hrankowski
Dec 24 '18 at 6:52
1
$begingroup$
That is exactly the point of the floor function. You round the fraction down to the next whole number. It is exactly what you asked for-the last of three is free.
$endgroup$
– Ross Millikan
Dec 24 '18 at 6:58
$begingroup$
Oh, you are right. I read your answer too quickly and didn't catch the notation. I'll delete my answer as it is therefore a duplicate.
$endgroup$
– Adam Hrankowski
Dec 24 '18 at 7:01
$begingroup$
Thanks a lot. It worked. I've edited my question to clarify X and N.
$endgroup$
– sdb1995
Dec 24 '18 at 7:04
$begingroup$
So the general formula would be if N >= X then we can pay for n - floor(n/x)
$endgroup$
– sdb1995
Dec 24 '18 at 7:10
add a comment |
$begingroup$
The simple answer is if one orders $n$ of the product you get $lfloor frac n3 rfloor$ free and pay for $n-lfloor frac n3 rfloor$
answered Dec 24 '18 at 6:46
Ross MillikanRoss Millikan
299k24200374
$endgroup$
$begingroup$
I don't think that will work. The discount must be on an integral number of widgets. There would only be one free widget for four purchased. Please see the example in the question, then see my answer.
$endgroup$
– Adam Hrankowski
Dec 24 '18 at 6:52
1
$begingroup$
That is exactly the point of the floor function. You round the fraction down to the next whole number. It is exactly what you asked for-the last of three is free.
$endgroup$
– Ross Millikan
Dec 24 '18 at 6:58
$begingroup$
Oh, you are right. I read your answer too quickly and didn't catch the notation. I'll delete my answer as it is therefore a duplicate.
$endgroup$
– Adam Hrankowski
Dec 24 '18 at 7:01
$begingroup$
Thanks a lot. It worked. I've edited my question to clarify X and N.
$endgroup$
– sdb1995
Dec 24 '18 at 7:04
$begingroup$
So the general formula would be if N >= X then we can pay for n - floor(n/x)
$endgroup$
– sdb1995
Dec 24 '18 at 7:10
add a comment |
$begingroup$
The simple answer is if one orders $n$ of the product you get $lfloor frac n3 rfloor$ free and pay for $n-lfloor frac n3 rfloor$
answered Dec 24 '18 at 6:46
Ross MillikanRoss Millikan
299k24200374
$endgroup$
The simple answer is if one orders $n$ of the product you get $lfloor frac n3 rfloor$ free and pay for $n-lfloor frac n3 rfloor$
answered Dec 24 '18 at 6:46
Ross MillikanRoss Millikan
299k24200374
answered Dec 24 '18 at 6:46
Ross MillikanRoss Millikan
299k24200374
answered Dec 24 '18 at 6:46
Ross MillikanRoss Millikan
299k24200374
answered Dec 24 '18 at 6:46
Ross MillikanRoss Millikan
299k24200374
299k24200374
$begingroup$
I don't think that will work. The discount must be on an integral number of widgets. There would only be one free widget for four purchased. Please see the example in the question, then see my answer.
$endgroup$
– Adam Hrankowski
Dec 24 '18 at 6:52
1
$begingroup$
That is exactly the point of the floor function. You round the fraction down to the next whole number. It is exactly what you asked for-the last of three is free.
$endgroup$
– Ross Millikan
Dec 24 '18 at 6:58
$begingroup$
Oh, you are right. I read your answer too quickly and didn't catch the notation. I'll delete my answer as it is therefore a duplicate.
$endgroup$
– Adam Hrankowski
Dec 24 '18 at 7:01
$begingroup$
Thanks a lot. It worked. I've edited my question to clarify X and N.
$endgroup$
– sdb1995
Dec 24 '18 at 7:04
$begingroup$
So the general formula would be if N >= X then we can pay for n - floor(n/x)
$endgroup$
– sdb1995
Dec 24 '18 at 7:10
add a comment |
$begingroup$
I don't think that will work. The discount must be on an integral number of widgets. There would only be one free widget for four purchased. Please see the example in the question, then see my answer.
$endgroup$
– Adam Hrankowski
Dec 24 '18 at 6:52
1
$begingroup$
That is exactly the point of the floor function. You round the fraction down to the next whole number. It is exactly what you asked for-the last of three is free.
$endgroup$
– Ross Millikan
Dec 24 '18 at 6:58
$begingroup$
Oh, you are right. I read your answer too quickly and didn't catch the notation. I'll delete my answer as it is therefore a duplicate.
$endgroup$
– Adam Hrankowski
Dec 24 '18 at 7:01
$begingroup$
Thanks a lot. It worked. I've edited my question to clarify X and N.
$endgroup$
– sdb1995
Dec 24 '18 at 7:04
$begingroup$
So the general formula would be if N >= X then we can pay for n - floor(n/x)
$endgroup$
– sdb1995
Dec 24 '18 at 7:10
$begingroup$
I don't think that will work. The discount must be on an integral number of widgets. There would only be one free widget for four purchased. Please see the example in the question, then see my answer.
$endgroup$
– Adam Hrankowski
Dec 24 '18 at 6:52
$begingroup$
I don't think that will work. The discount must be on an integral number of widgets. There would only be one free widget for four purchased. Please see the example in the question, then see my answer.
$endgroup$
– Adam Hrankowski
Dec 24 '18 at 6:52
1
1
$begingroup$
That is exactly the point of the floor function. You round the fraction down to the next whole number. It is exactly what you asked for-the last of three is free.
$endgroup$
– Ross Millikan
Dec 24 '18 at 6:58
$begingroup$
That is exactly the point of the floor function. You round the fraction down to the next whole number. It is exactly what you asked for-the last of three is free.
$endgroup$
– Ross Millikan
Dec 24 '18 at 6:58
$begingroup$
Oh, you are right. I read your answer too quickly and didn't catch the notation. I'll delete my answer as it is therefore a duplicate.
$endgroup$
– Adam Hrankowski
Dec 24 '18 at 7:01
$begingroup$
Oh, you are right. I read your answer too quickly and didn't catch the notation. I'll delete my answer as it is therefore a duplicate.
$endgroup$
– Adam Hrankowski
Dec 24 '18 at 7:01
$begingroup$
Thanks a lot. It worked. I've edited my question to clarify X and N.
$endgroup$
– sdb1995
Dec 24 '18 at 7:04
$begingroup$
Thanks a lot. It worked. I've edited my question to clarify X and N.
$endgroup$
– sdb1995
Dec 24 '18 at 7:04
$begingroup$
So the general formula would be if N >= X then we can pay for n - floor(n/x)
$endgroup$
– sdb1995
Dec 24 '18 at 7:10
$begingroup$
So the general formula would be if N >= X then we can pay for n - floor(n/x)
$endgroup$
– sdb1995
Dec 24 '18 at 7:10
add a comment |
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