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# M13-16

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Channarong22 wrote:
Bunuel : Why we minimize C=10???

We have that a + b = 16. Since a and b are distinct integers, then a = b = 8 is not possible, so the least value of b is 9 (7 + 9 = 16). Hence the least value of c is 10.
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Hi Bunuel
Just a thought
I follow your logic and get that $$d=14$$ and $$c = 10$$, and I answered correctly. But I do notice some edge cases here

However, if we consider that $$a+b=2*8=16$$, the distinct possibilities for the solutions for $$a + b$$ are:
$$[1, 15]$$
$$[2, 14]$$
.
.
.
$$[7, 9]$$

if we consider, for example, $$a = 1, b = 15$$, which satisfy $$a + b = 16$$, then $$c$$ MUST BE greater than 15. I.E, $$c = 10$$, the premise of If the average of the smaller two of these four integers is 8 breaks down, since now the two smallest integers are $$a = 1, c = 10$$, and the average IS NOT 8.

Wouldn't it be better if the question stem explicitly say $$a < b < c < d$$? I know it not absolutely necessary, just to cross out said edge case

Thanks

Bunuel wrote:
Official Solution:

The average (arithmetic mean) of four distinct positive integers is 10. If the average of the smaller two of these four integers is 8, which of the following represents the maximum possible value of the largest integer?

A. 12
B. 14
C. 15
D. 16
E. 17

Assume the four distinct positive integers are $$a, b, c$$, and $$d$$, such that $$0 < a < b < c < d$$.

Given that the average of these integers is 10, we know that $$a+b+c+d=4*10=40$$;

Furthermore, we're told that the average of the smallest two of these integers is 8, which means that $$a+b=2*8=16$$. So, $$16+c+d=40$$, and thus $$c+d=24$$.

To find the maximum possible value for $$d$$, we need to minimize $$c$$. The smallest value of $$c$$ can be 10, for $$a=7$$ and $$b=9$$ (this is because $$a + b = 16$$ and since $$a$$ and $$b$$ must be distinct integers, it's impossible for both to be 8. Thus, the smallest possible value for $$b$$ is 9). So, with the smallest possible value for $$c$$ as 10, we have $$10+d=24$$ which gives us $$d=14$$.

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r_putra_rp wrote:
Hi Bunuel
Just a thought
I follow your logic and get that $$d=14$$ and $$c = 10$$, and I answered correctly. But I do notice some edge cases here

However, if we consider that $$a+b=2*8=16$$, the distinct possibilities for the solutions for $$a + b$$ are:
$$[1, 15]$$
$$[2, 14]$$
.
.
.
$$[7, 9]$$

if we consider, for example, $$a = 1, b = 15$$, which satisfy $$a + b = 16$$, then $$c$$ MUST BE greater than 15. I.E, $$c = 10$$, the premise of If the average of the smaller two of these four integers is 8 breaks down, since now the two smallest integers are $$a = 1, c = 10$$, and the average IS NOT 8.

Wouldn't it be better if the question stem explicitly say $$a < b < c < d$$? I know it not absolutely necessary, just to cross out said edge case

Thanks

Bunuel wrote:
Official Solution:

The average (arithmetic mean) of four distinct positive integers is 10. If the average of the smaller two of these four integers is 8, which of the following represents the maximum possible value of the largest integer?

A. 12
B. 14
C. 15
D. 16
E. 17

Assume the four distinct positive integers are $$a, b, c$$, and $$d$$, such that $$0 < a < b < c < d$$.

Given that the average of these integers is 10, we know that $$a+b+c+d=4*10=40$$;

Furthermore, we're told that the average of the smallest two of these integers is 8, which means that $$a+b=2*8=16$$. So, $$16+c+d=40$$, and thus $$c+d=24$$.

To find the maximum possible value for $$d$$, we need to minimize $$c$$. The smallest value of $$c$$ can be 10, for $$a=7$$ and $$b=9$$ (this is because $$a + b = 16$$ and since $$a$$ and $$b$$ must be distinct integers, it's impossible for both to be 8. Thus, the smallest possible value for $$b$$ is 9). So, with the smallest possible value for $$c$$ as 10, we have $$10+d=24$$ which gives us $$d=14$$.

All is correct both in the question and the solution. To maximize d, we need to to minimize c, for that we need to minimize b. The minimum value of b such that 0 < a < b and a + b = 16, is 9 (since all are integers).
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