The following partially-filled table shows the marks scored by 5 students named A, B, C, D and E in five different subjects, namely P, C, M, B and S. The marks scored by all students in all subjects were natural numbers.

         P   C   M   B   S  Total
    A        87  72      58 340
    B        64      96     354
    C                       318
    D                       402
    E    67  44             388
Total   338 332 383 398 351 

the following points are known:

  1. There is exactly one student who scored 99 marks in one subject, which was the highest score among all students in all subjects.

  2. There is exactly one student who scored 44 marks in one subject, which was the lowest score among all students in all subjects.

    1. C got 3 different perfect square marks in 3 different subjects and equal non-perfect square marks in M and S.
    2. D scored 2 marks more in C than in P and 3 marks more in M than in S.
    3. C’s highest score was in P.
    4. B’s highest score was 96 in B but D got the highest score in B, which was a prime number.
    5. Except in M, all scores of D were prime numbers. B got equal marks in M and S.

The highest marks scored by any student in any subject were scored in which of the following subjects?

Cannot be determined

Please share your approach for this.

Source : Me

  • $\begingroup$ My first thought was "what does 'S' stand for?" I see my question is answered in the text! (at (3)) $\endgroup$ – Jonathan Allan Oct 20 '18 at 17:50
  • $\begingroup$ What do u mean when you say "Source : Me". Is this is your own question or from someone else? $\endgroup$ – Kevin L Oct 22 '18 at 6:59
  • $\begingroup$ It is my question. $\endgroup$ – sam Oct 22 '18 at 7:59

Edit-note: all of this hinges on the assumption that each subject represents the appropriate initial.

I believe the answer is


We know that C's scores are

49, 64, 81, leaving 124 for the other two equal scores; they're both 62. Since C's highest score is in P, that's 81; M and S are both given to be 62.

We also know that D's highest score in B has to be

97, the only prime number between 96 and 99.

Now to figure out the rest of the scores in C:

We know C's score there is either 49 or 64. Because we're missing 137 total there and D's score is known to be a prime, it's not 49 (that would leave D with 88), so it's 64 (meaning D scored a 73). This also fills in C's score for B, which is 49, and D's score for P, which is 73-2 = 71.

Following up we can complete D's scores too:

We're short 161 points on the line. Since M = S+3, we can solve for both by subtracting 3 and dividing by 2: S = 158/2 = 79 and M = 79+3 = 82.

Now, it would seem we have no more info, but

we know A can't have the 99 because his total scores are short 123, and there's no score below 44; we know B can't have it either because his highest score is 96. Thus, E must have our missing score of 99.

Now, considering B's M and S scores are known equal:

by looking for the remainder for the M and S columns, we're missing 167 and 152 points respectively; since the only other missing scores are B's equal ones, we know E's scores for those subjects are X+15 and X.


We're missing a total of 277 points in E's scores. If we assume that his score for B is 99, we would have to fill in the remaining 178 using 2x+15 points, which is impossible through only natural numbers.


the 99 was scored in Mathematics, by E.

  • $\begingroup$ Your answer is correct. Thanks a lot for your effort. $\endgroup$ – sam Oct 20 '18 at 10:36

My answer is

The highest score is 99 in Math.


6. Tells me that D's Biol must be 97 because that is the only prime between B's 96 and 99.
3. Tells me that C got 49, 64, 81 in Phys, Chem, Biol (in some order). That leaves 124 total of Math and Stat, 62 each.
5. Tells me that C's 81 was in Phys, leaving Chem and Biol with 49 or 64 each.
4. and 7. Tell me that D's Phys and Chem must be either 59 and 61, or 71 and 73 (the only primes in the range different by 2). Considered with the possible 49 or 64 in the row above, C and D's marks for Chem must be 64 and 73 to get the correct sum. So now I know that C's Biol is 49 and D's Phys is 71.
4. Now tells me that D's Math + Stat must be 161 and with a difference of 3, one a prime, those marks must be 82 and 79.

Thus far:

    P   C   M   B   S   Tot
A       87  72      58  340
B 64 96 354
C 81 64 62 49 62 318
D 71 73 82 97 79 402
E 67 44 388
Tot 338 332 383 398 351
Where is the 99?
It can't be A's because that will give the other subject a mark less than 45.
It can't be B's because it would leave the sum of the two equal marks as an odd number.
So it must be E's, either Math or Stat (97 was the highest Biol). If the 99 is in E's Stat the table can be completed by subtractions leaving A's Phys with only 31.
Therefore E's Math is 99.


    P   C   M   B   S   Tot
A   61  87  72  62  58  340
B 58 64 68 96 68 354
C 81 64 62 49 62 318
D 71 73 82 97 79 402
E 67 44 99 94 84 388
Tot 338 332 383 398 351

  • 1
    $\begingroup$ A belated answer because I had already figured it out, but was struggling with the "hidden" format. If only there was one simple button! $\endgroup$ – Weather Vane Oct 20 '18 at 11:11
  • $\begingroup$ Thanks for answering with your approach. It doesnt matter even if u posted later. Approach and answer should be correct. $\endgroup$ – sam Oct 20 '18 at 11:42
  • $\begingroup$ Weather Vane would you try and answer puzzling.stackexchange.com/questions/73855/water-island-puzzle/…? with your approach if possible $\endgroup$ – sam Oct 20 '18 at 11:57
  • 1
    $\begingroup$ It can't be B's because it would leave the sum of the two equal marks as an odd number. - simpler: It can't be B's because "B’s highest score was 96 in B" :) $\endgroup$ – Jonathan Allan Oct 20 '18 at 18:46
  • $\begingroup$ @JonathanAllan good spot thank you. $\endgroup$ – Weather Vane Oct 20 '18 at 18:47

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