+1713 Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Laila's score on the last test?
+4622 Let's try to break this problem down.
If the average score Laila got was 82, then her total score was 82*5=410 points.
We can label the tests that she took as x, x, x, x, y. The x's are the tests that she got the same score on, and y is the score that she got a different score on.
Now, let's try some casework.
Since she received a higher score on the last test, let's pick a value, and maybe one-hundred(100)? If she has score 100 on the last test, then she must have gotten 4x+100=410, 4x=310, x=77.5 points on the tests she had the same score on. Yet, this does not work, because she scored an integer number of points on all five tests.
Thus, our goal is when we subtract the value from 410, the result should be divisible by four. If y=98, and if she had scored ninety-eight points on the last test, we have her scoring seventy-eight(78) points on the remaining four tests. This is one possibility.
If 98 works, then 94 should work, as this also yields a multiple of four. If she has scored 94 points on the last test, then she would have scored 79 points on the remaining tests,
Trying some of them, y=90 point works, then x=80 points. Also, y=86 points works, then x=81 points. However, if y=82 points, then x=82 points. We do not want this, such as given in the problem. Thus, the only possible values of Laila scoring the number of points on the last test, which is denoted as (y) is, y=98, y=94, y=90, and y=86. Thus, only four(\(\boxed{4}\)) possibilities work.
Hope this helps,
-tertre
Let's denote x as the value of the score for the first 4 tests. Also, the 5th test score can be denoted as y. The total score would be 5 * 82 = 410.
We have a Diophantine equation that is:
\[4x + y = 410 \]
One possible value of y is 98 which means x is 78. We can decrease y by 4 and increase x by 1 until x is equal to or larger than y to find other possible values since the overall change is 0.
\[(4 * 78) + 98 = 410 \]
\[(4 * 79) + 94 = 410 \]
\[(4 * 80) + 90 = 410 \]
\[(4 * 81) + 86 = 410 \]
This results in 4 possible values for the last test.
