Show the value is possible
The set
1, 2, 3, 4, 5, 6, 7, 108, 764
contains nine distinct positive integers and totals 900. Therefore 764 is achievable.
Chapter 3 · Lesson 3.5
Extreme Values under Average Constraints
平均数条件下的最大值与最小值
An average fixes the total like a budget. To make one value as large or as small as possible, move every other value to the edge allowed by the conditions—then prove that no better result can exist.
Fix the total. Respect every condition. Pack the other values to the edge.
By the end of this lesson, I can…
- turn an average into a fixed total;
- identify lower bounds, distinctness, fixed values, and ranking conditions;
- maximize one value by minimizing the other allowed values;
- minimize a ranked value by packing the surrounding values as tightly as possible;
- prove an extreme value with both a construction and an impossibility argument;
- recognize when the information does not determine one answer.
Fixed total = limited budget
What the other values use is no longer available to the value you are trying to maximize. All given averages in this lesson are exact.
Mission 1
Not completed
The total is a fixed budget
When the count and average are fixed, the total is fixed. Think of that total as a budget shared by all the values.
Total = average × count
Five values have average 7
Their total must always be 5 × 7 = 35. Move the slider to change how much of the total belongs to the highlighted value.
Target 17
Others 18
Fixed total35
Target value17
Other values total18
Mission 2
Not completed
Read every constraint before optimizing
The words around the numbers decide which values are allowed. Missing one condition can completely change the answer.
Smallest-companion machine
The machine finds the four smallest legal companion values. “Reserved value to skip” means that value is forbidden in these four slots, whether or not repetition is allowed. The questions below use their stated conditions, independent of your machine settings.
Smallest legal companions
Positive, distinct, and 3 is reserved: use 1, 2, 4, 5. Their sum is 12.
1Lower bound
Positive integers start at 1. Nonnegative integers also allow 0.
2Distinctness
“Different” or “distinct” means no repeated values.
3Fixed and ranked values
Reserve values already given, and preserve the required order.
Mission 3
Not completed
Push the companion values to the edge
Start with a smaller model: five distinct positive integers have a total of 35.
Make the largest value as large as possible
Use the four smallest legal companions. Whatever remains from 35 becomes the largest value.
Largest value—
Companion sum—
Fixed total35
Remaining target—
Reverse question: To make the largest value as small as possible, first order each candidate set from smallest to largest, then compare its last value. Pack all five distinct integers tightly. The set 5, 6, 7, 8, 9 totals 35, so 9 is achievable. A largest value of 8 would allow at most 4 + 5 + 6 + 7 + 8 = 30, which is too small.
Mission 4
Not completed
A largest-possible value from a fixed average
A set of distinct positive integers has average 100. One value is 108. After removing 108, the remaining average is 99.
Count recap: 108 is 8 above the old average of 100. Removing those 8 extra points makes the remaining average fall by 1, so 8 values remain. There were 9 values originally.
Original total = 9 × 100 = 900
Maximize one unknown value
Besides the fixed value 108 and the target value, seven companion values remain. To maximize the target, make those seven companions as small as the conditions allow.
Fixed108
Target—
Seven companions—
Fixed value108
Target—
Mission 5
Not completed
An extreme answer needs a proof
Finding a large-looking value is not enough. A complete extreme-value argument has two parts.
Show one step larger is impossible
If the target were 765, the seven companions would have only:
900 − 108 − 765 = 27
But seven distinct positive integers require at least 1 + 2 + ··· + 7 = 28. Therefore 765 is impossible. Every integer above 765 leaves an even smaller companion budget, so all values larger than 764 are impossible.
Mission 6
Not completed
Minimize a ranked score
Six students have distinct integer scores with average 92.5. The highest score is 99 and the lowest is 76. What is the smallest possible third-highest score? Order the six scores from highest to lowest; the third-highest is the third score in that order.
Total score = 6 × 92.5 = 555
Maximum total allowed by a candidate
To test a low third-place candidate, make every other score as large as the ranking permits:
Candidate94
Largest possible total552
Even the largest possible total is 552, so a third-highest score of 94 is too low.
The second-highest can be at most 98. If the third-highest is t, the fourth and fifth can be at most t − 1 and t − 2.
Threshold: With third place 95, the packed scores 99, 98, 95, 94, 93, 76 total exactly 555. With third place 94, even the maximum possible total is only 552. Therefore the minimum is 95.
Mission 7
Not completed
Is there enough information for one answer?
Eight distinct positive integers total 45. After removing the smallest and largest, the remaining six total 33. What is the smallest of the remaining six?
Smallest extreme + largest extreme = 45 − 33 = 12
That condition does not force one unique middle set. Reveal two valid examples.
Valid Set A
1234591011
Middle-six total: 2 + 3 + 4 + 5 + 9 + 10 = 33. The smallest remaining value is 2.
Valid Set B
234567810
Middle-six total: 3 + 4 + 5 + 6 + 7 + 8 = 33. The smallest remaining value is 3.
Mathematical conclusion: as written, the conditions allow more than one answer. A trustworthy solver checks uniqueness instead of forcing a single number.
Mission 8
Not completed
Error detective
Each student tried to create an extreme value but broke one condition or skipped part of the proof.
ARepeated companions
For five distinct positive integers totaling 35, Mina uses 1, 1, 1, 1 to maximize the last value.
BForbidden zero
For distinct positive integers, Kai starts the companion list with 0.
CNo barrier check
Lena claims 765 is possible in the 900-total problem without checking the seven-companion budget.
DBroken ranking
A list says the third-highest score is 94 but the fourth-highest is 95.
Mission 9
Not completed
Extreme-value workshop
For each problem, write the fixed total, list the constraints, and decide which other values should be pushed upward or downward.
1Maximum largest
Four distinct positive integers total 30. What is the largest possible value?
Hint
Reserve 1, 2, and 3 for the other values.
2Zero is allowed
Five distinct nonnegative integers total 40. What is the largest possible value?
Hint
The four smallest companions are 0, 1, 2, and 3.
3One value is fixed
Six distinct positive integers average 15. One value is 20. What is the largest possible value of another number?
Hint
Total 90. Besides 20 and the target, four companion slots remain.
4Minimum largest
Five distinct positive integers total 35. What is the smallest possible largest value?
Hint
Pack five consecutive values as closely as possible.
5Minimum second place
Five distinct integer scores average 80. The highest is 90 and the lowest is 70. What is the smallest possible second-highest score?
Hint
Find the total of the three middle scores. For a proposed second-place score, how large could their combined total be?
6Maximum smallest
Four distinct positive integers total 30. What is the largest possible smallest value?
Hint
Try four consecutive values. If the smallest were one larger, what minimum total would be forced?
7Find Liu Jia's score
Liu Jia joins four students scoring 78, 79, 82, and 91. Liu Jia's score is 6 points above the five-person average. What is Liu Jia's score?
Hint
If Liu Jia scores x, the group average is x − 6.
8Find the rank
Using the score from Question 7, what is Liu Jia's rank from highest to lowest?
Hint
Place the score among 91, 82, 79, and 78.
Worked proof: minimum second place (Question 5)
The total is 5 × 80 = 400. Removing 90 and 70 leaves 240 for the three middle scores. If second place were at most 80, those scores could total at most 80 + 79 + 78 = 237, too little. Second place 81 works: 90, 81, 80, 79, 70 total 400. Therefore the minimum is 81.
Worked proof: maximum smallest value (Question 6)
If the smallest were at least 7, four distinct integers would total at least 7 + 8 + 9 + 10 = 34, exceeding 30. A smallest value of 6 works: 6 + 7 + 8 + 9 = 30. Therefore 6 is the largest possible smallest value.
Mission 10
Not completed
Exit ticket
Complete these five questions without reopening the worked examples.
Keep checking the constraints
An extreme value must fit the fixed total and every condition.
Certificate of Mastery
Extreme-Value Strategist
This certifies that
can convert averages to fixed totals, honor positivity, distinctness, fixed-value and ranking constraints, construct extreme examples, and prove that no more extreme value is possible.
Teaching notes
The lesson retains the original's conditions—distinct positive integers, average 100, and a fixed value of 108—and develops the maximum 764 through both construction and impossibility.As written, that information does not determine a unique answer: the page therefore uses it as a constraint-sufficiency investigation and displays two valid counterexamples rather than silently assigning one answer.All interactive models, added examples, feedback, workshop questions, and the objective exit ticket are instructional scaffolds created to make the lesson self-contained.