Dan's data drops down

We find that $\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}=\dfrac{153}{140}$ so any of the fractions could be the incorrect one, and so we don't get anywhere here.

Now, the problem becomes finding a number, 700-750, that is a multiple of four elements of ${3,4,5,6,7}$ .

We must take at least one multiple of 2, and one multiple of 3. Then, if we take 7, the multiple of 42 in the range is 714, which is $42\cdot 17$ . This is not a multiple of either 4 or 5, so we cannot take 7.

Then, our number is a multiple of 3,4,5,6, making it a multiple of 60, and therefore 720.

We find the answer is $720-240-180-144-120=\boxed{36}$ .

Let $x$ be the number of students such that $700 \leq x \leq 750$ .

We know that exactly one is wrong and the others are right. As such $x$ must(at least) be a multiple of $LCM(a,b,c,d,e)$ which lies in the above mentioned range.

The LCM of the right fractions depending upon which is wrong are:

• $LCM(3,4,5,6)=60$
• $LCM(3,4,5,7)=420$
• $LCM(3,4,6,7)=84$
• $LCM(3,5,6,7)=210$
• $LCM(4,5,6,7)=420$

It is clear that for $x$ to lie in our range only the first case from above is possible, i.e the Level $5$ fraction is wrong. As such if there exists an answer to the question then $x=720$ is the only possible value. We easily verify that this is possible.

Therefore, number of Level $5$ students on Brilliant $= 720 \cdot(1-{\displaystyle \sum \limits_{i=3}^6 \frac{1}{i}} )=720 \cdot \frac{1}{20}=\boxed{36}$

Note that since the number of students is a whole number, it has to be divisible by all the denominators of the fractions in their simplest forms.

Since there is exactly one wrong fraction, we first proceed to find the least common multiple (lcm) of different quadruples of the denominators.

$lcm(4,5,6,7)=420, lcm(3,5,6,7)=210, lcm(3,4,6,7)=84$

$lcm(3,4,5,7)=420, lcm(3,4,5,6)=60$

Among these numbers, only 60 divides a number between 700 to 750 (720). Hence the total number of students is 720 and the number of level 5 students is

$720(1-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6})=36$ .

Let $X$ denote the number of Brillant students, so $700 \leq X \leq 750$ . We have exactly four statements below that are true and one is false.

• $X$ is divisible by 3

• $X$ is divisible by 4

• $X$ is divisible by 5

• $X$ is divisible by 6

• $X$ is divisible by 7

Suppose the first statement is not true, then $X$ is a multiple of $lcm (4,5,6,7) = 420$ , there's no satisfies range of $X$ given above that is a multiple of $420$ , so the first statement must be true.

Similarly, suppose the second statement is not true, then $X$ is a multiple of $lcm (3,5,6,7) = 210$ , there's no satisfies range of $X$ given above that is a multiple of $210$ , so the first statement must be true.

And, suppose the third statement is not true, then $X$ is a multiple of $lcm (3,4,6,7) = 84$ , there's no satisfies range of $X$ given above that is a multiple of $84$ , so the first statement must be true.

And, suppose the fourth statement is not true, then $X$ is a multiple of $lcm (3,4,5,7) = 420$ , there's no satisfies range of $X$ given above that is a multiple of $420$ , so the first statement must be true.

Lastly, suppose the fifth statement is not true, then $X$ is a multiple of $lcm (3,4,5,6) = 60$ , given the satisfied range of $X$ above, $X = 720$ only, so the number of level 5 students is $720 (1 - \frac {1}{3} - \frac {1}{4} - \frac {1}{5} - \frac {1}{6} ) = 36$

Hence the fifth statement is false, so our answer is $\boxed{36}$

Since it is given that exactly one of Dan's fractions is incorrect, we know that exactly $5-1=4$ of his fractions are correct. We also know that since these fractions are correct, the number of Brilliant students must be divisible by each of the denominators. (For example, if $\frac{1}{3}$ is correct, then we know that the number of students is divisible by 3.) Now, we can test the possible cases of which fraction is incorrect.

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Case 1: divisible by 3, 4, 5, 6, and 7

Though one of these fractions is incorrect, it is still possible for the number to be divisible by all five numbers 3, 4, 5, 6, and 7. However, the least common multiple of the five numbers 3, 4, 5, 6, and 7 is equal to $3\times2^2\times5\times7=420$ , and no integer in the range 700-750 is divisible by 420, so we know that this is not the case.

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Case 2: Divisible by 4, 5, 6, and 7, but not 3

We notice that since 3 divides 6, if a number is not divisible by 3, then it must not be divisible by 6 either. Because of this, this case is not possible.

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Case 3: Divisible by 3, 5, 6, and 7, but not 4

This means that the fraction that is incorrect is $\frac{1}{4}$ , which means that the other four fractions are correct. However, in order for that to happen, we need to have a multiple of the least common multiple of 3, 5, 6, and 7 be in the range or 700-750. The least common multiple of 3, 5, 6, and 7 is equal to $2\times3\times5\times7=210$ . However, there is no integer in the range 700-750 that is divisible by 210, so we know that this case is not true.

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Case 4: Divisible by 3, 4, 6, and 7, but not 5

Proceeding like before, we take the least common multiple of 3, 4, 6, and 7. We find this to be $2^2\times3\times7=84$ . However, $84\times8=672$ and $84\times9=756$ , so there is not integer in the range 700-750 that is divisible by 84. This is not the case.

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Case 5: Divisible by 3, 4, 5, and 7, but not 6

Notice that if a number is divisible by 3 and 4, then it must be divisible by the least common multiple of 3 and 4, which is 12. However, since 6 divides 12, the number must be divisible by 6 as well, so we know that this is not the case.

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Case 6: Divisible by 3, 4, 5, and 6, but not 7

The least common multiple of 3, 4, 5, and 6 is equal to $2^2\times3\times5=60$ . This time, there is an integer in the range 700-750 that is divisible by 60, and that number is $60\times12=720$ . Because of this, we know that this is the case we want.

We know that the fraction $\frac{1}{7}$ is the incorrect fraction because 720 is not divisible by 7. To find the correct fraction in place of that, we have

$1-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}=\frac{1}{20}$ .

We know this works, because the question states that every user is in Level 1, 2, 3, 4, or 5, and only one of these. Solving for the number of students in Level 5, we have $\frac{1}{20}\times720=\boxed{36}$

if 1/3 is wrong then 1/4+1/5+1/6+1/7 = x/420. this means no. of children should be a multiple of 420, but we see none of the multiples of 420 lie b/w 700-750.so this is not the case.

if 1/4 is wrong , then repeating the process(1/3+1/5+1/6+1/7) we get : y/210 . But again none of the multiples of 210 lies b/w 700-750.

Continuing this process till the case of 1/6 , none of our assumptions is wrong...

THUS 1/7 is wrong. Now 1/3+1/4+1/5+1/6 = 57/60 .

This means no. of students is a multiple of 60 and multiple of 60 that lies b/w 700-750 is 720. Thus no. of students = 720.

Now 3/60 of them are 5th level students [ since 1-57/60 = 3/60].

thus 5th level students= 720 * 3/60

So ANSWER => 36<=

• LCM (3,4,5,6) = 60
• LCM (3,4,5,7) = 420
• LCM (3,4,6,7) = 84
• LCM (3,5,6,7) = 210
• LCM (4,5,6,7) = 420

Now, there are no multiples of 420, 84 or 210 in the given range.

So, it is not possible with these combinations of denominators

So, our correct fractions by Dan must be: $\frac{1}{3} \frac{1}{4} \frac{1}{5} \frac{1}{6}$

$\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} = \frac{19}{20}$

so, the other fraction for level 5 is $\frac{1}{20}$

Also, the only multiple of 60 (LCM (3,4,5,6)) in the given range is 720 . So that should be the exact no. of Brilliant students

So, in level 5 we have $\frac{1}{20} \times 720 = \boxed{36}$

The question means that the number of students is exactly divisible by 3,4,5,6,7 with one of them being wrong. We can observe that divisibility by 3 cannot be wrong as the number is divisible by 4 and 6 too.The number should also be divisible by 4,6 and 7(if none of them are wrong) but there is no number between 700 and 750 satisfying.

$\Rightarrow$ One of 4,6 and 7 is wrong.Only 7 can be wrong as the number is divisible by 3 and 5.

So, our number is divisible by 3,4,5 and 6 which comes out to be $720$ .

Number of students in level 5 = $720-(240+180+144+120) = \boxed{36}$

the number of all students must have all the denominator of the correct ratios as a divisor.

there are only two numbers in the range from 700 to 750 which share 5 and 7 as a divisor, they are 700 and 735

but 3 and 6 are not a divisor of 700 so it can't be the number of the students

and 4 and 6 are not a divisor of 735 so it can't be the number of the students

so the wrong ratio is one of the 1/5 or 1/7

since 3 4 7=84, 700/84=8.3, and 750/84=8.9 so there is no number in that range share 3, 4, 6, and 7 as a divisors so 1/7 is the wrong ratio.

since 750/(3 4 5)=12.5, then the number of students is 12 3 4*5=720

fifth level students are = (1-1/3-1/4-1/5-1/6)*720=36

Suppose that 1/7 is uncorrect. So what number from 700 to 750 is divisible by 3-4-5-6? Only 720. 720/3+720/4+720/5+720/6+x=720 684+x=720 => x=36. 720/36=20. The correct fraction of 5 level student is 1/20.

The sum of the 5 fractions is 1+13/140. The wrong denominator must satisfy: 1/x -13/140 = 1/y where y is the correct integer. The unique solution is x=7, y=20. Next, the least common multiple of all denominators is 3x4x5 = 60. Number of students must be a multiple of 60. Given the costraints, it is 720. Finally, level 5 students are 720/20 = 36.

Since only one of the fractions is wrong the rest should give integer values on dividing the total number of students..

It can be seen that only 720 can be divided by four (3,4,5,6) of the factors (3,4,5,6,7). Thus it is the total number of students..

Finding the number of students of L5 (i.e of the wrong factor 7) ::

Total students(1-Sum of the true fractions)

720 [1-(1/3)-(1/4)-(1/5)-(1/6)]

ANS:: 36

Since the number is divisible by 5, then the units value of total number of Brilliant students might be 0 or 5. Let $6!$ = $6*5*4*3*2*1$ = $720$ to satisfying divisors of $3$ , $4$ , $5$ and $6$ . $720/3 = 240$ $720/4 = 180$ $720/5 = 144$ $720/6 = 120$ The only wrong prediction is $1/7$ for Level 5 Brilliant students. Therefore, $720 - (240 + 180 + 144 + 120)$ = $36$

〔3,4,5,6,7〕＝420→:(

〔4,5,6,7〕＝420→:( ←420×1.714≈700 ＆ 420×1.786≈750

〔3,5,6,7〕＝210→:( ←210×3.429≈700 ＆ 210×3.571≈750

〔3,4,6,7〕＝420→:( ←420×1.714≈700 ＆ 420×1.786≈750

〔3,4,5,6〕＝60→: ) ←60×11.67≈700 ＆ 60×12.5≈750

60×12＝720 720×⅓＝240 720×¼＝180 720×1/5＝144 720×1/6＝120 (720-240-180-144-120)/720＝36/720＝4/80＝1/20

700×1/20＝36

1/3 +1/4 + 1/5 +1/6. = 0. It is known that the overall should be lesser than one. So 1/20 which is 0.5 will be an exact solution. This would mean the answer is 35(700/20) but 37 is also a possible solution for 750 students. Therfore the answer would be 36