There's So Many Selection!
You're in a stationary shop where a pen cost 50p and a pencil cost 15p. You want to buy a collection of pens and pencils and you have £25 (£1 is equivalent to 100p) to spend.
In how many ways can you spend exactly £25 pounds on buying those pens and pencils?
Details and Assumptions :
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One way of spending exactly £25 pounds in that shop is buying only pens, that this spending £25 pounds on buying 50 pens. It is up to your personal choice on how many pens and how many pencils you want to buy, but before making a decision, you need to consider all possible combinations of set of pens and pencils you can buy, exactly in your budget.
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You have no restriction on the minimum number of pens or pencils you should buy.
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Another example of spending exactly £25 pounds on stationary collection is buying 20 pens and 100 pencils.
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1 solution
Let X = number of pencils, Y= number of pens
We need to find integer solutions for:
15X + 50Y = 2500
One integer solution is X=0 and Y= 50
To find all integer solutions, we use Linear Diophantine equations which says:
The general solution for ax + by = d, where d is the multiple of gcd(a,b) is:
X= x + k(b/g), Y= y - k(a/g),
where g = gcd(a,b), (gcd= greatest common divisor) and x, y is one of the integer solutions to the equation:
aX + bY= d.
gcd(15,50) = 5, x = 0, y = 50
So all integer solutions to the equation is given by:
X= k(50/5), Y= 50 -k(15/5)
X= 10k, Y= 50-3k, where k is an integer
To find the range of values of k, we have the condition that X>=0 and Y>=0 as number of pens or pencils cannot be negative.
It follows that:
10k >= 0, 50-3k >= 0
0 =< k =< 50/3
50/3= 16.66666667
Therefore, counting the number of values k can take from 0 to 16 (as k should be an integer) gives 17 solutions for k. The values k can take are: 0, 1, 2, 3, ........, 16.
The general solution is X = 10k, Y= 50- 3k and it has 17 integer solutions for X and Y which satisfy
15X + 50Y = 2500.
So, there are 17 possible combinations of number of those pens and pencils you can get in exactly £25.
One example:
At k= 5,
X= 10(5), Y= 50- 3(5)
X= 50, Y= 35
This combination is 50 pencils and 35 pens :)