Comparing the number votes obtained by A across different constituencies, the lowest number of votes were in constituency

Party	S1	S2	S3	S4	S5	Total
A						49
B						35
C						16
Total	20	20	20	20	20	100
Winner	A	C	C	B	B

How many votes does a winner need in one constituency?

• Suppose the winner got only 7. The other two parties would share the remaining 13, so at least one of them would also have 7 or more, and there would be no clear winner
• So the winner of any constituency needs at least 8 votes

C wins both S2 and S3, so C has at least 8 in each. That is already 16, which is exactly C's total, so there is nothing left over for the other three seats.

Party	S1	S2	S3	S4	S5	Total
A						49
B						35
C	0	8	8	0	0	16
Total	20	20	20	20	20	100
Winner	A	C	C	B	B

With C out of the picture in S1, S4 and S5, those three seats are straight A vs B fights over all 20 votes, so the winner there needs at least 11.

• S1 goes to A, so B gets at most 9 there
• S4 goes to B, so B has at least 11
• S5 goes to B, so B has at least 11, but B's votes are strictly increasing, so B in S5 must beat B in S4

"Number of votes obtained by B in S1, S2, S3, S4 and S5 are distinct natural numbers in increasing order."

That pushes B in S5 to at least 12.

Party	S1	S2	S3	S4	S5	Total
A						49
B	≤ 9			≥ 11	≥ 12	35
C	0	8	8	0	0	16
Total	20	20	20	20	20	100
Winner	A	C	C	B	B

Now S2 and S3, where C takes 8 of the 20 votes

• A and B share the remaining 12
• Neither of them can reach 8, otherwise C would not be the winner
• Two numbers adding to 12 with both below 8 must each be 5, 6 or 7

So B's votes in S2 and S3 both come from {5, 6, 7}, and being increasing, they add to at least 5 + 6 = 11.

Party	S1	S2	S3	S4	S5	Total
A						49
B	≤ 9	5-7	5-7	≥ 11	≥ 12	35
C	0	8	8	0	0	16
Total	20	20	20	20	20	100
Winner	A	C	C	B	B

Squeezing B's total of 35 from both sides

S4 and S5 already take at least 11 + 12 = 23, which leaves the first three seats at most 12:

B(S1) + B(S2) + B(S3) ≤ 35 − 23 = 12

Going the other way, B(S2) + B(S3) is at least 11 and B(S1) is a natural number, so it is at least 1. Those three add to at least 12.

Both bounds land on 12, so every value is forced:

• B(S1) = 1
• B(S2) + B(S3) = 11, and from {5, 6, 7} in increasing order that is 5 and 6
• B(S4) + B(S5) = 35 − 12 = 23, and with 11 and 12 as minimums that is exactly 11 and 12

Party	S1	S2	S3	S4	S5	Total
A						49
B	1	5	6	11	12	35
C	0	8	8	0	0	16
Total	20	20	20	20	20	100
Winner	A	C	C	B	B

Each column holds 20 votes, so A's row is simply whatever is left after B and C:

• S1: 20 − 1 − 0 = 19
• S2: 20 − 5 − 8 = 7
• S3: 20 − 6 − 8 = 6
• S4: 20 − 11 − 0 = 9
• S5: 20 − 12 − 0 = 8

Party	S1	S2	S3	S4	S5	Total
A	19	7	6	9	8	49
B	1	5	6	11	12	35
C	0	8	8	0	0	16
Total	20	20	20	20	20	100
Winner	A	C	C	B	B

A's votes come to 19 + 7 + 6 + 9 + 8 = 49, exactly as given.

Checking the winners once more: A takes S1 with 19, C takes S2 and S3 with 8 each, and B takes S4 and S5 with 11 and 12. Every condition is satisfied, so this grid is the unique arrangement and all questions can be read off it.

A's row reads 19, 7, 6, 9, 8, so A's votes were lowest in S3.