any reason why it shouldn't be 329 ?

Merlinaeus wrote: any reason why it shouldn't be 329 ? Because all the fractions are stated as being less than the true value.

Yes. But 110/329 > 1/3 99/329>3/10 120/329>4/11 And 110 + 99 + 120 = 329

But, $\frac{58}{173} > \frac{1}{3}$ $\frac{52}{173} > \frac{3}{10}$ $\frac{63}{173} > \frac{4}{11}$

We shall prove that the construction that xeroxia provided is the answer. Suppose $a$ students go to the cinema, $b$ students go to the theatre, and $c$ students go to the concert, and the total number of students is $S = a+b+c$. We wish to find $a,b,c \in \mathbb{Z}$ such that $3a > S , 10b > 3S , 11c > 4S$ all hold simultaneously. Suppose $2a = b+c+k, k=1$. We shall later consider the possibilities when $ k > 1 $. The previous inequalities are equivalent to \[ 2a > b + c \] \[ 7b > 3a+3c \] \[ 7c > 3b + 3c \] Generally, if $2a=b+c+k$, then $7b > \frac{3}{2}(b+c+k) + 3c \implies 11b > 3k + 9c,$ and $7c > 2(b+c+k) + 4b \implies 5c>6b+2k.$ Combining these, we have that $11b > 3k + \frac{9}{5}(6b+2k) \implies b > 33k$, where $k$ is some positive integer. Using $2a = b+c+1$, we establish upper and lower bounds on $c$ in terms of $b,k$. From the 2nd inequality, we get that \[7b > \frac{3}{2} (b+c+1) + 3c \implies 14b >3b+9c+3 \implies \frac{11b-3}{9} > c \] Similarly, from the 3rd inequality, we have that \[ 7c > 2(b+c+1) + 4b \implies 5c > 6b+2 \implies c > \frac{6b+2}{5} \] Hence, we need the interval \[ \left ( \frac{6b+2}{5} , \frac{11b-3}{9} \right)\] to contain an integer. It suffices to consider the endpoints of the interval modulo 1 (that is, the fractional part of each number). If the (nonzero) fractional part of the right endpoint is less than the (nonzero) fractional part of the left endpoint, that means that an integer must be contained in the interval. We can then consider the equivalent, reduced interval to be \[\left ( \frac{b + 2}{5} , \frac{2b-3}{9} \right)\] We now utilize the substitution $ b' = b-33 $ for ease of calculations. We know that $b \geq 34$, and hence $b' \geq 1$. This changes the interval to \[ \left( \frac{b'+35}{5} , \frac{2b' + 63}{9} \right) \equiv \left( \frac{b'}{5} , \frac{2b'}{9} \right), \] where we subtracted 7 from both expressions since they are unnecessary. To make it simpler to compare the sizes of these fractions, using a common denominator helps. Hence, we are left with \[ \left( \frac{9b'}{45} , \frac{10b'}{45} \right) \], and wish the find the smallest $b'$ such that the fractional part of the left endpoint is greater than the fractional part of the right endpoint. We begin computations from $b' =1$ (denominators are omitted for visibility). \[(9,10), (18,20), (27,30), (36, 40), (0,5), (9,15)\] \[(18,25), (27,35), (36,0), (0,10), (9,20), (18,30)\] \[(27,40), \boxed{(36,5)}, (0,15), (9,25), (18,35),(27,0), \boxed{(36,10)}\] The first boxed pair corresponds to $b' = 14, b=47$ and the interval $ \left( \frac{284}{5}, \frac{514}{9} \right)$. This interval obviously contains $57$, which would be our value for $c$. However, this yields $2a = b+c+1 = 105$, implying $a$ is not an integer, which is impossible. However, the second boxed pair corresponds to $b' = 19, b=52$ and the interval $ \left( \frac{314}{5}, \frac{569}{9} \right)$, which does contain the integer $63$. Now, $2a = 116 \implies a = 58$, and so the minimum sum when $k=1$ is $19+52+58 = 173$. If $k \geq 2$, then $b > 66$, and so it is not possible for $k\geq2$ to yield a triplet $(a,b,c)$ with a sum less than what we have already constructed. Hence, the minimum number of possible students is $\boxed{173}$.

Suppose there are n students. WLOG, we suppose all students either are going to cinema, theatre or concert (if somebody doesn't, let them go to any place and the condition still hold, but with more students - contradict to the "minimize"). Students go to cinema: >= bottom(n/3) + 1 Students go to theatre : >= bottom(3n/10) + 1 Students go to concert : >= bottom(4n/11) + 1 Considering the "minimize", we have bottom(n/3) + 1 + bottom(3n/10) + 1 + bottom(4n/11) + 1 = n Let n = 3a + b, here b>=0 and b <= 2, then: n = 3a + b = 3 + bottom((3a+b)/3) + bottom((9a+3b)/10) + bottom((12a+4b)/11) = 3 + 3a + bottom((3b-a)/10) + bottom((a+4b)/11) That is, b - 3 = bottom((3b-a)/10) + bottom((a+4b)/11) Let a+4b = 11c + d, here d>=0 and d<=10, we have b - 3 = bottom((7b-11c-d)/10) + c = bottom((7b-c-d)/10) Check b by case: Case 1, b = 0, -3 = bottom(-(c+d)/10) ==> c+d > 20, a = 11c+d >= 11 X 11 + 10 = 131 n >= 393 Check: bottom(393/3) + 1 = 132, bottom(3*393/10) + 1 = 118, bottom(4*393/11) + 1 = 143, 132+118+143=393 Case 2, b = 1, -2 = bottom(-(c+d-7)/10) ==> c+d-7 > 10 or c+d >= 18, a = 11c+d - 4>= 11*8+10 - 4= 94, n >= 3*94 + 1 = 283 Check: 283/3 = 94 1/3, 3*283/10 = 84 9/10, 4*283/11 = 102 10/11, 95 + 85 + 103 = 283 Case 3, b = 2, -1 = bottom(-(c+d-14)/10) ==> c+d-14 > 0 or c+d >= 15, a = 11c+d - 8 >= 11*5+10 - 8 = 57, n >= 3*57 + 2 = 173 Check: 173/3 = 57 2/3, 3*173/10 = 51 9/10, 4*173/11 = 62 10/11, 58 + 52 + 63 = 173 Finally, the answer is 173.

Answer is 173