Lights on!

A floor will be seen by all its factors.

Between 1-50 , there are:

$1$ number which has $1$ factor

$15$ numbers having $2$ factors

$4$ numbers having $3$ factors,

$15$ numbers having $4$ factors

$1$ number having $5$ factors

$8$ numbers having $6$ factors ,

No number having $7$ factors.

$4$ numbers having $8$ factors,

$1$ number each having $9$ and $10$ factors .

If all 50 floors are switched on , we check that $\displaystyle \sum_{i = 1}^{50} x_{i} = 207.$

Hence we now have to switch off the lights of floors in such a manner that sum of their factors is $7$ .

Case 1 : 49 floors have lights on :

Now we should switch off a floor having 7 factors , which is impossible Hence no such way.

Case 2 : 48 floors have lights on :

Now we have to switch off two floors such that the sum of their factors is 7. (i) The floors switched off have factors $6$ and $1$ , ${8 \choose 1} {1 \choose 1}$ = $8$ ways.

(ii)The floors switched off have factors $5$ and $2$ , ${1 \choose 1} {15 \choose 1}$ = $15$ ways.

(iii)The floors switched off have factors $4$ and $3$ , ${15 \choose 1} {4 \choose 1}$ = $60$ ways.

Case 2 : Now we have to switch off 3 floors such that the sum of their factors is 7.

(i) Floors switched off have factors $1$ , $2$ and $4$ , ${1 \choose 1}{15 \choose 1}{15 \choose 1} = 225$ ways.

(ii) Floors switched off have factors $2$ , $2$ and $3$ , ${15 \choose 2} {4 \choose 1} = 420$ ways.

(iii) Floors switched off have factors $3$ , $3$ and $1$ , ${4 \choose 2}{1 \choose 1} = 6$ ways.

Hence, total no. of ways = $8$ + $15$ + $60$ + $225$ + $420$ + $6$ $= \fbox{734}$ .

First it is useful to calculate what the sum would equal if the lights on all floors were turned on. This is given by $\sum_{i = 1}^{50} \lfloor \frac {50}{i} \rfloor$ , which is actually quite easy to brute force and find $207$ .

Next, I claim that the number of people who look at a certain floor is given by the number of divisors of that floor number. This is because a person with number $i$ sees a floor iff the floor number is divisible by $i$ . Say a person with an $i$ that is not a divisor of the floor number sees that floor. Since that person only sees floors that are a multiple of $i$ , then that means that floor number must be divisible by $i$ , which is a contradiction.

Since at least $47$ floors had their lights on, we look to find combinations of $1, 2,$ or $3$ positive integers $f \leq 50$ such that the $\sum σ_0(f) = 7$ ( $σ_0(f)$ is the divisor function). Let us consider these cases separately:

If there is only $1$ floor which has their lights off, then the number of divisors of that floor number must be $7$ . Since any number with more than one unique prime factor yields a composite number of divisors, the number we are looking for must be a prime sixth power. The lowest such number is $2^6 = 64$ , which is not less than $50$ , so there are no such floors.

When there are two floors with their lights off, the number of divisors can be split as follows: $(1, 6), (2, 5), (3, 4)$ . So now we need to find the number of integers less than 50 with such a number of divisors.

There is only one integer with $1$ divisor, and that is $1$ . A number with two divisors is prime, and there are $15$ primes less than 50. For an integer to have 3 divisors it must be a prime power of two (using the same logic as above, since 3 is prime), which yields $4, 9, 25, 49$ . For an integer to have 4 divisors it must either be a prime power of three or a number which can be expressed as $p * q$ where $p$ and $q$ are primes. This yields $8, 27$ and $2 * 3 \rightarrow 2*23, 3*5 \rightarrow 3*13, 5*7 = 2 + 13 = 15$ integers. For five divisors it must be a prime power of four leaving only $16$ as an answer, and finally for six divisors it must either be $2^5 = 32$ or a number of the form $p^2 * q$ , which yields another $1 + 7$ solutions. To recap, $1 \rightarrow 1, 2 \rightarrow 15, 3 \rightarrow 4, 4 \rightarrow 15, 5 \rightarrow 1, 6 \rightarrow 8$ .

Now that we found these, we can easily calculate the number for each case, namely $8 + 15 + 60 = 83$

When 3 floors have their lights off, then the following figurations can be conceived: $(2, 2, 3), (1, 3, 3), (1, 2, 4)$ . Note that there can only be a single 1 in any configuration since 1 is the only number with 1 divisor. Using the values we calculated before, we get the number of answers in this case to be $4{15 \choose 2} + {4 \choose 2} + (15)(15) = 420 + 6 + 225 = 651$ .

Thus, the total number of configurations is $651 + 83 = 734$ .

First let us investigate the case when all $50$ floors have their lights on. Then, the $i^{th}$ man records the floors which are multiples of $i$ . Thus, $x_i$ will simply be the number of multiples of $i$ in the range $[1,50]$ , i.e $x_i= \left \lfloor \frac{50}{i} \right \rfloor$

Then, $\sum_{i=1}^{200} x_i= \sum_{i=1}^{200} \left \lfloor \frac{50}{i} \right \rfloor$ We can manually compute the sum and show that it is equal to $207$ . Thus all $50$ floors cannot have their lights switched on. Let $n$ floors have their lights switched off, and let $\{a_1, a_2, \cdots , a_n \}$ be the set of the number of floors which have their lights switched off. Each $a_i$ will be missed by exactly $d(a_i)$ people (where $d(X)$ denotes the number of positive divisors of $X$ ). We want exactly $7$ floors to be missed, so we must have $\sum_{i=1}^{n} d(a_i)= 207-200= 7$ Before we proceed on doing casework on $n$ , let's first find the number of solutions to $d(m)=n$ for $1 \leq n \leq 7$ and $1\leq m \leq 50$ .

$\underline{d(m)=7}$ For $d(m)$ to be $7$ , we must have $m=p^6$ for some prime $p$ . The minimum possible value of $m$ will then be $2^6=64$ , which is clearly greater than $50$ . Hence this equation does not have any solutions.

$\underline{d(m)=6}$ For $d(m)$ to be $6$ , $m$ must be either of the form $p^5$ or $pq^2$ for some distinct primes $p$ and $q$ . If $m=p^5$ , the only possibility of $p$ which keeps $m$ in the desired range is $2$ . Let's investigate the second case. For $p=2$ , the possibilities for $q$ are $3, 5$ . For $p=3$ , the only possibility of $q$ is $2$ . For $p=5$ , the only possibilities of $q$ are $2$ and $3$ . For $p=7$ and $p=11$ , we have $q=2$ . There are no possibilities for $q$ when $p>11$ . Thus the equation $d(m)=6$ has $8$ solutions.

$\underline{d(m)=5}$ For $d(m)$ to be $5$ , $m$ must be equal to $p^4$ for some prime $p$ . The only possibility for $p$ is $2$ , so $d(m)=5$ has $1$ solution.

$\underline{d(m)=4}$ For $d(m)$ to be $4$ , $m$ has to be either equal to $p^3$ or $pq$ for some distinct primes $p$ and $q$ . If $m=p^3$ , the only possibilities for $p$ are $2$ and $3$ . Let's see the second case. For $p=2$ , the possibilities for $q$ are $3,5,7,11,13,15,17,19,23$ . Proceeding similarly, one can show that the second case has $13$ solutions. So $d(m)=4$ has $15$ solutions in total.

$\underline{d(m)=3}$ For $d(m)$ to be $3$ , $m$ has to be of the form $p^2$ . The possibilities for $p$ are $2,3,5,7$ , so this equation has $4$ solutions in total.

$\underline{d(m)=2}$ For $d(m$ to be $2$ , $m$ has to be a prime. Since there are $15$ primes $\leq 50$ , this equation has $15$ solutions in total.

$\underline{d(m)=1}$ The only possible value of $m$ is $1$ , so this equation has only $1$ solution.

Now we are ready to do casework on the number of floors which had their lights off.

Exactly $1$ floor had its lights off
Let $x$ be the number of the floor which had its switch off. Then we must have $d(x)=7$ . But by our previous analysis, this case has no solutions.

Exactly $2$ floors had their lights off
Let $x$ and $y$ ( $x \neq y$ ) be the number of the floors which had their lights off. Then we have $d(X)+d(Y)=7$ . By symmetry assume $d(X) \geq d(Y)$ . Below we tabulate the possibilities for $(d(X),d(Y))$ and its number of solutions.
Here is the link in case the picture doesn't load
This gives $60+75+8= 83$ solutions in total.

Exactly $3$ floors had their lights off
Let $x$ , $y$ , and $z$ (pairwise distinct) be the numbers of the floors which had their lights off. Then we have $d(X)+d(Y)+d(Z)=7$ . By symmetry assume $d(X) \geq d(Y) \geq d(Z)$ . The possibilities for $(d(X),d(Y),d(Z))$ are $(3,2,2)$ , $(3,3,1)$ , $(4,2,1)$ , and $(5,1,1)$ . By similar casework (most of which is omitted here because of its length), we can show that this equation has ${651}$ solutions in total.

So the total number of possible combinations is $651+83= \boxed{734}$ .

Note In this process, we have proved the following interesting lemma which can be easily generalized to $n$ in place of $50$ : $\sum_{i=1}^{n} d(i)= \sum_{i=1}^{n} \left \lfloor \frac{n}{i} \right \rfloor$ Also note that we were basically finding the number of solutions to $d(X)+d(Y)=7$ and $d(X)+d(Y)+d(Z)=7$ . Was the time $7$ p.m a hint to indicate this? :)

Let f(n) be the number of divisors of n. (Also known as the sigma 0 function). For each floor, n, f(n) people will check it. Thus it being off subtracts f(n) from the maximal count. Let g(k) be the number of integers, n, less than 50 for which f(n) =k. Some tedious calculation yields: G(1)=1 G(2)=15 G(3)=4 G(4)=15 G(5)=1 G(6)=8 G(7)=0 G(8)=4 G(9)=1 G(10)=1 The maximum possible sum of x i becomes 1 g(1)+2 g(2) ...+10 g(10)=207 Calvin counted 200. Therefore if floors a, b, and c are off then f(a)+f(b)+f(c)=7. Keeping in mind that at most 3 floors had their lights off, we partition 7 into at most 3 parts. Then, for each partition, calculate the number of ways it could occur. This is the number of ways you could choose 3 or fewer floors to have their lights off and still have a sum of 200. 7,6+1,5+2,4+3,5+1+1,4+1+2,3+2+2, and 3+3+1 can occur in 0,8,15,60,0,225,420, and 6 ways, respectively. Adding them all together yields the correct answer, 734.