What other deductions are possible?
In a class of n students, a test was administered and only scored on whole numbers from 0-10 if every possible score was scored at least once, the average score was seven.
What can you deduce?
Deduction 0: the average of the first 11 is five.
1.There are at least 11 students
55+x=7*(11+z)
z is the number of students beyond the first eleven.
x=22+7*z.
x/z=average score of the additional beyond 11 is an integer.
x/z=22/z+7.
If z=2, z=11, z=22 this equation is satisfied.
Case one: x/2=11+7=18- a contradiction. Case two: x/11=9 Case Three: x/22=1+7=8
Deduction two: on average the number of students beyond the first 11 scored on average a nine or a eight.
Deduction Three: You either n=22 or n= 33 .
If you have 22 students in total, then the average of the last 11 is nine.
If you have 33 students total then the average of the last 22 is eight.
How do you use these four deductions to find the lowest score that could have been scored by more than one student or exactly one student?
You are correct to say that there must be more than $11$ students, since there are $11$ different marks and the simple average of those is $5$, which is less than $7$.
But your other deductions do not look so good. There is no requirement for the "average score of the additional beyond $11$ is an integer."
You can say there must be at least $n\ge 19$ students, since the total number of marks is at most $5 \times 11 +10 \times (n-11) = 10n-55$, which is less than $7n$ when $n<\frac{55}{3} \approx 18.33$.
Having $19$ students is possible since $\frac{1+2+3+4+5+6+7+8+9+10+10+10+10+10+10+10+10+8}{19}= \frac{133}{19}=7$.
Any greater number of students is also possible, for example with more scoring $7$, though other patterns are also possible.