Royston had d boxes. He put 7 doughnuts into each of the boxes. Then he had 3 doughnuts left. What is the original number of doughnuts Royston had?
Let's use algebra to solve the problem.
Let's say the original number of doughnuts Royston had was "x".
He put 7 doughnuts into each of the d boxes, so he used a total of 7d doughnuts.
He also had 3 doughnuts left over.
So we can set up the equation:
x = 7d + 3
But we don't know the value of "d", the number of boxes.
However, we know that "d" must be a whole number, since you can't have a fraction of a box.
We also know that when you divide any whole number by 7, the remainder must be either 0, 1, 2, 3, 4, 5, or 6.
So if we divide the left side of the equation (x) by 7, the remainder must be 3, since Royston had 3 doughnuts left.
Therefore:
x รท 7 = q remainder 3
where "q" is a whole number (the quotient).
We can rewrite this equation as:
x = 7q + 3
Now we can substitute this into the original equation:
7q + 3 = 7d + 3
We can simplify:
7q = 7d
q = d
So "q" and "d" are just two different names for the same number.
Therefore, the original number of doughnuts Royston had was:
x = 7d + 3 = 7q + 3
where "q" is any whole number.
This means that there are infinitely many possible answers to this question. For example:
- If Royston had 10 boxes, then he originally had 73 doughnuts (7 times 10 plus 3).
- If Royston had 20 boxes, then he originally had 143 doughnuts (7 times 20 plus 3).
- If Royston had 100 boxes, then he originally had 703 doughnuts (7 times 100 plus 3).
And so on.
To find the original number of doughnuts Royston had, we need to subtract the 3 leftover doughnuts from the total number of doughnuts distributed into the boxes.
Let's assume the original number of doughnuts Royston had is x.
He put 7 doughnuts into each of the d boxes, so he distributed a total of 7d doughnuts.
Since he had 3 doughnuts left, we can write the equation:
7d + 3 = x
Therefore, the original number of doughnuts Royston had is x = 7d + 3.