Mean Median Mode Range 438938

1. **State the problem:** We have a collection of eight integers where the mean, median, unique mode, and range are all equal to $8$. We want to find the largest integer in this collection. 2. **Define variables and known facts:** Let the eight integers be $a_1 \leq a_2 \leq a_3 \leq a_4 \leq a_5 \leq a_6 \leq a_7 \leq a_8$. - Mean $= 8$ means $$\frac{a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8}{8} = 8 \implies a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 = 64.$$ - Median $= 8$ means the average of the middle two numbers is 8: $$\frac{a_4 + a_5}{2} = 8 \implies a_4 + a_5 = 16.$$ - Unique mode $= 8$ means $8$ appears more times than any other number, and only $8$ is the mode. - Range $= 8$ means $$a_8 - a_1 = 8.$$ 3. **Analyze median condition:** Since $a_4 + a_5 = 16$ and $a_4 \leq a_5$, the simplest choice is $a_4 = 8$ and $a_5 = 8$. 4. **Mode condition:** $8$ must appear more times than any other number. Since $a_4$ and $a_5$ are both $8$, there are at least two $8$s. 5. **Range condition:** $a_8 - a_1 = 8$. 6. **Sum condition:** Total sum is 64. 7. **Goal:** Maximize $a_8$. 8. **Strategy:** To maximize $a_8$, minimize $a_1$ because $a_8 = a_1 + 8$. 9. **Try $a_1 = x$, then $a_8 = x + 8$:** Sum of all numbers is 64. 10. **Assign known values:** - $a_4 = 8$, $a_5 = 8$. 11. **Mode 8 must be unique:** So $8$ must appear more times than any other number. 12. **Try to have as many 8s as possible to ensure mode:** Let's try four 8s: $a_3 = 8$, $a_4 = 8$, $a_5 = 8$, $a_6 = 8$. 13. **Now the numbers are:** - $a_1 = x$ - $a_2 = y$ (with $x \leq y \leq 8$) - $a_3 = 8$ - $a_4 = 8$ - $a_5 = 8$ - $a_6 = 8$ - $a_7 = z$ (with $8 \leq z \leq a_8 = x + 8$) - $a_8 = x + 8$ 14. **Sum equation:** $$x + y + 8 + 8 + 8 + 8 + z + (x + 8) = 64$$ $$2x + y + z + 40 = 64$$ $$2x + y + z = 24.$$ 15. **Constraints:** - $x \leq y \leq 8$ - $8 \leq z \leq x + 8$ - Mode is 8, so no other number appears 4 or more times. 16. **Try to minimize $x$ to maximize $a_8 = x + 8$:** - Try $x = 0$ (lowest integer): Then $a_8 = 8$. Sum equation: $2(0) + y + z = 24 \implies y + z = 24$. Since $y \leq 8$ and $z \geq 8$, max $y$ is 8, so $z = 16$. Check if $z \leq a_8 = 8$? No, $16 \leq 8$ is false. So $x=0$ invalid. 17. **Try $x=1$:** $a_8 = 9$. Sum: $2(1) + y + z = 24 \implies y + z = 22$. $y \leq 8$, so max $y=8$, then $z=14$. Check $z \leq a_8=9$? No. Invalid. 18. **Try $x=2$:** $a_8=10$. Sum: $4 + y + z = 24 \implies y + z = 20$. Max $y=8$, so $z=12$. Check $z \leq 10$? No. Invalid. 19. **Try $x=3$:** $a_8=11$. Sum: $6 + y + z = 24 \implies y + z = 18$. Max $y=8$, so $z=10$. Check $z \leq 11$? Yes. Also $z \geq 8$. So $z=10$, $y=8$. 20. **Check mode uniqueness:** - Numbers: $3, 8, 8, 8, 8, 8, 10, 11$ (we must assign $a_2 = y = 8$, $a_7 = z = 10$) Count of 8s: 5 times. Others appear once or twice. Mode is unique and 8. 21. **Check median:** $a_4=8$, $a_5=8$, median is 8. 22. **Check range:** $a_8 - a_1 = 11 - 3 = 8$. 23. **Check sum:** $3 + 8 + 8 + 8 + 8 + 8 + 10 + 11 = 64$. 24. **All conditions satisfied.** 25. **Try $x=4$ to see if larger max possible:** $a_8=12$. Sum: $8 + y + z = 24 \implies y + z = 16$. Max $y=8$, so $z=8$. Check $z \leq 12$ and $z \geq 8$ valid. Numbers: $4, 8, 8, 8, 8, 8, 8, 12$. Count of 8s: 6 times. Mode unique. Sum: $4 + 8 + 8 + 8 + 8 + 8 + 8 + 12 = 64$. Range: $12 - 4 = 8$. Median: $a_4=8$, $a_5=8$. All conditions met. 26. **Try $x=5$:** $a_8=13$. Sum: $10 + y + z = 24 \implies y + z = 14$. Max $y=8$, so $z=6$. But $z \geq 8$ (since $a_7 \geq a_6=8$), invalid. 27. **So max $a_8$ is 12 when $x=4$.** **Final answer:** The largest integer in the collection can be $\boxed{12}$.