As Simple as 3-5-7

We can infer that the $(N-3-3)$ numbers in between are $\geq \frac{5}{3}$ and $\leq \frac{7}{3}$ . Thus, the largest possible number achievable by the $N$ numbers can be obtained by $(N-3-3) \times \frac{7}{3} + 5 + 7$ and the smallest possible number formed can be obtained by $(N-3-3) \times \frac{5}{3} + 5 + 7$

$Case 1$ : $N = 9$ .We infer that the 3 numbers in the middle $\leq 7$ as the sum of the 3 largest numbers are $7$ , we arrive that the largest number we can achieve is $5+7+7 = \boxed{19}$ which is $< 20$ . Thus $N \neq 9$

$Case 2$ : $N = 10$ . The largest possible number we can achieve is $(10-3-3) \times \frac{7}{3} + 5 + 7= 9\frac{1}{3} + 5 + 7 = \boxed{21\frac{1}{3}}$ . Since the answer is $> 20$ , we now check for the smallest possible number which is $(10-3-3) \times \frac{5}{3} + 5 + 7= 6\frac{2}{3} + 5 + 7 = \boxed{19\frac{2}{3}}$ . Since $20$ is within the range of the smallest possible number and the largest possible number, it is possible that $N=10$ .

$Case 3$ : $N = 11$ . To save time, we will only work on the smallest possible number as the largest possible number is definitely $>20$ . The smallest possible number that can be obtained is $(11-3-3) \times \frac{5}{3} + 5 + 7= 8\frac{1}{3} + 5 + 7 = \boxed{20\frac{1}{3}}$ which is $>20$ . Thus, $N \neq 11$

The smallest possible value of the third "smallest" number is 5/3, and the largest possible value of the third "largest" number is 7/3. (I put in the quotation marks to acknowledge the "not necessarily unique" condition).

Now the remaining (N - 6) numbers must add to 8. Since 5 * (5/3) = 25/3 > 8 we know that (N - 6) < 5. Next, since 3 * (7/3) = 7 < 8 we know that (N - 6) > 3.

We are thus left with N - 6 = 4, i.e., N = 10, and an example of 10 numbers meeting the conditions is 1,2,2,2,2,2,2,2,2,3.

Thus the answer is "II only".

The sum of those mid numbers is 8. So the sum of each blocks is 5-8-7 respectively. The averages of each block must be in ascending or descending order. First Block= 5/3 = 1.67 Second Block: 8/3=2.67 X 8/4=2 / 8/5=1.6 X Third Block = 7/3= 2.33

So this will be obtained with this sequence of averages: 1.67,2,2.33 So the mid block must have 4 numbers. Hence, 10 only :)

1 2 2 ...............2 2 3

rest nos sum up to 8

all nos are 2 so 4 more nos

total 10 nos

We will show that the smallest numbers must contain a $2$ , and the largest numbers must contain a $2$ . Hence, the middle sequence must contain all $2$ s, for a total sequence length of $10$ ( Case II ).

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The Three Smallest Numbers

The sum of the three smallest numbers must be $5$ , as given. Denote the largest of these as $x$ . We will show $x \geq 2$ . Suppose:

$x \leq 0$ : Then, all three numbers are non-positive and the sum is at most $0 < 5$ . So, $x > 0$ .

$x = 1$ : The maximum sum is $1 + 1 + 1 = 3 < 5$ . So, $x > 1$ .

$x = 2$ : For example, $1 + 2 + 2 = 5$ . Hence, $x \geq 2$ .

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The Three Largest Numbers

The sum of the three largest numbers must be $7$ , as given. Denote the smallest of these as $y$ . Because $x \geq 2$ , then $y \geq 2$ . We will show $y = 2$ . Suppose:

$y = 2$ . For example, $2 + 2 + 3 = 7.$ So, $y \geq 2$ .

$y = 3$ . Then, the smallest possible sum is $3 + 3 + 3 = 9 > 7$ . Any larger $y$ will result in a larger sum. So, $y = 2$ .

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Because $y = 2$ , the largest numbers must contain a $2$ . Because $x \geq 2$ and $y = 2$ , the smallest numbers must contain a $2$ as well. Hence, the middle sequence must solely contain four $2$ s ( $5 + 4*2 + 7 = 20$ ), for a total length of $10$ ( Case II ).

We know through logical reasoning that none of the numbers are negative. So, three numbers in the start (a,b,c) and three in the end (x,y,z). An unknown amount of numbers in the middle. We know A+B+C=5 and X+Y+Z=7. 5+7=12. We know that the sum of all the numbers is 20. So the numbers in the middle must add up to 8 (20-12).

There are a number of possibilities for the first 3 numbers.

a) 0,1,4 (3 numbers greater than or equal to 4 cant add up to 7)

b) 0,2,3 (3 numbers greater than or equal to 4 cant add up to 7)(I said 4 because 4+4 is the only way to get 8 in the middle)

c) 1,1,3 (Same argument as option (b))

d) 1,2,2 (2+2+2+2 forms 8 in the middle)(2+2+3 forms 7 in the end)

(D) is the only possible answer. Hence, 3 in the start, 3 in the end, 4 two's in the middle.

Answer : N can ONLY be 10.

First three number 1,2,2 and last three will be 2,2,3 ......so rest is 4 numbers with total sum of 8. 12,2,2,2,2,2,2,2,3