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Question 1208021: On dart board, a player can only score 2, 6 or 9 points for a throw. What is the smallest number of throws needed to get a total score of exactly 82 points?
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52788)   (Show Source):
You can put this solution on YOUR website! .
The idea of solution is to use "9 points" throws as many times as possible
in order for to gain the rest scores using "2 points" and "6 points" throws.
As soon as you get this idea, you will find the answer in one minute (or faster):
eight "9 points" throws (that will give you 72 scores), and
one "6 points" and two "2 points" throws (to get the rest 10 scores).
In all, you need 8 + 1 + 2 = 11 throws. ANSWER
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Solved.
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In the United States, most public school students learn multiplication table in second or third grade.
(This info is from the Internet).
This problem is good to give it to the students of the 2nd or 3rd grade,
immediately after they learned the multiplication table.
It is a right way to develop their minds in a timely manner.
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
x = number of 2 pointers
y = number of 6 pointers
z = number of 9 pointers
Each variable is some nonnegative whole number {0,1,2,3,...}
2x+6y+9z = total number of points
2x+6y+9z = 82
x+y+z = number of throws
The goal is to minimize x+y+z
Since z is attached to the largest coefficient, we'll make z as large as possible to find the minimum of x+y+z.
If we made x as large as possible, then we'd be at the other end of the spectrum.
82/9 = 9.111 approximately
This rounds down to 9.
If z = 9, then 9z = 9*9 = 81
That leaves 82-81 = 1 point left, but of course there's no way to score 1 point with 2 pointers and/or 6 pointers.
So there's no way to get exactly 82 points with z = 9.
We'll decrease z by 1.
If z = 8, then 9z = 9*8 = 72
There would be 82-72 = 10 points left, and now it appears to be possible to achieve a score of 10 with 2 pointers and/or 6 pointers.
The subproblem is now 2x+6y = 10
This turns into x+3y = 5 after dividing everything by 2.
Use a bit of trial-and-error, or you can use the Extended Euclidean Algorithm, to find one solution is (x,y) = (2,1)
Sure enough 2x+6y = 2*2+6*1 = 4+6 = 10
What are some other integer solutions on x+3y = 5?
Solve for y to get y = (-1/3)x+5/3.
It has slope -1/3 which means to go from one point to another we move down 1 and right 3.
Starting at (2,1) follow the "down 1, right 3" movement pattern to arrive at (5,0)
Go back to (2,1). Follow the reverse of that movement pattern to go up 1, left 3, and you'll arrive at (-1,2)
However, recall that x cannot be negative.
The only practical nonnegative integer solutions to x+3y = 5 are (2,1) and (5,0)
In the first case we have x+y+z = 2+1+8 = 11 shots
In the second case we have x+y+z = 5+0+8 = 13 shots
The first case is the winner in terms of fewest number of throws.
Answer: 11
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