Combinatorics
A class of 10 students took a math test. Each problem was solved by exactly 7 of the students. If the first nine students each solved 4 problems, how many problems did the tenth student solve?
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2 solutions
Let there be x problems. So total problems are 7x. So the tenth solved
7
x
−
3
6
which should be an integer. So least x is 6.
Check:-
7
∗
6
=
4
2
.
T
o
t
a
l
s
o
l
v
e
d
=
4
∗
9
+
6
=
a
l
s
o
4
2
6
.
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At the heart, this is a "construction problem". You not only need to find necessary conditions (to find the numerical answer), you also need to demonstrate existance (for a complete solution.)
By "check", I meant "find a configuration which satisfies the conditions". We should construct an example of 10 students and 6 questions, specifying which questions these students answered correctly, and showing that it can be achieved.
In the example that I gave, your check will be
7 × 2 = 1 4 . Total solved = 1 × 9 + 5 = 1 4 .
However, the problem comes about from "there are 2 questions, but the last person answered 5 correctly". This is the sanity check that Ankit performed, in terms of saying that 3 6 + n ≥ 7 n , which in this case will be 9 + x ≥ 7 x , which is not true for x = 2 .
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Thank you. I will keep this in mind. After check had I put:-
7
∗
x
=
4
∗
9
+
x
would it been OK?
Not quite enough.
If instead the first nine students each solved 1 problem (instead of 4), then your approach will yield 7 x − 9 , and so x is at least 5. However, there is no scenario where we can have 10 students which satisfy the conditions.
You should also check that this situation can indeed be achieved.
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Thanks. Because the way Latex operate, the word check was misplaced. I should have corrected it. Is it OK now ?
Just see for multiples of 7.you will get the answer
The total number of correct answers must be a multiple of 7. The first 9 people got 36 = 1 mod 7 correct answers in all. So the 10th person solved some number of problems equal to 6 mod 7. We also know that if the 10th person solved n problems, then 36+n ≥ 7n because there are at least n problems and 7 people answered each one. So the 10th person solved 6 problems.