Mean Girls Math Problems And Solutions

The movie Mean Girls has 3 interesting math questions. This video presents the problems and their solutions.

Solutions
First problem: 1:28
Second problem: 2:57
Third problem (limit): 4:01

Power series approximation idea

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[JLvatron]

The last question is very relevant to the movie on a whole, because it proves there is no limit to how mean those Mean Girls could be!

[skatelife]

if you dont know the power series approximations, simply use l'Hopital's rule:

if f(x) and g(x) are differentiable functions and d/dx (g(x)) does not equal to 0, then

lim x->0 f(x)/g(x) = lim x-> f'(x)/g'(x)

you stop when you dont get the undefined 0/0 or infinity/infinity, which is -2/0 as well, so its not defined.

[cga]

Great video,
I haven't seen the movie, but the way the second problem is redacted, 741 would be counted as correct because they are asking for 1 of the possible answers (It says AN odd three digit number, not THE odd three...), and btw, 345 and 147 also meet the criteria

[verlepte]

About the solution to the second problem: it says find an odd number that satisfies these conditions. It does not say it has to be a unique solution, so both 741 and 543 are correct, as are 345 and 147.

[earth]

For the second one, you can start with 444 and change the digits.

For example you can add 1 to the hundredths and subtract one from the ones, and get 543.

As long as you are adding and removing the same amount, and not changing the middle digit, this will work.

[HemmligtNavn]

couldn't you just use L'Hospitals rule for the 3rd question?

[DanielWalvin]

3:35 "not enough information"

There is, because it says "find *an* odd three digit number" -- any of the four possibilities is correct.

[peorakef]

3:40 so what. it says "find one" ad not "find the". no problem at all.

[pault]

The second question says "find AN odd number. . .", not find THE odd number. It doesn't matter that more than one number works, they just had to produce one. That's assuming the remainder of the cut-off question didn't impose further restrictions than assumed.

[paulflute]

yes you could figure out the second problem from what you heard in the movie.. In fact you did .. Twice.. the problem does not ask for a 'unique' number.. just a number that satisfies.. so any will do..

[delanomighty]

For the second question, note that they asked for ~a~ number, not a specific one. Based on that, any of the following four numbers could work: 147, 345, 543, 741.

[witzar]

Ad 2. Let m be the middle digit, s=(m-d) the smaller and l=(m+d) the larger. We have (m-d) + m + (m+d) = 12, so m=4. Digit s has to be odd (otherwise all digits are even and the number is even), so either s=1 (then l=7) or s=3 (then l=5). This gives us all possible solutions: 147, 741, 345, 543.

[christiandinkel]

Problem 2 seems solvable with this information, since the problem statement reads "find an odd..." and not "find the od...", so both answers seem correct.

[hasanjakir]

The second problem has enough information to be solved.
Suppose,
a+(a+b)+(a+2b)=12
or, 3(a+b)=12
or, a+b=4
So, a=4 - b

Here, a has to be an odd integerand b has to be positive integer.
So, a€{1, 3, 5, 7, 9}
b€{1, 2, 3, 4, 5, 6, 7, 8, 9, 0}
Now, use the equation a=4 - b
The solitions are:
{a, b}={1, 3}, {3, 1}
So, the digits are 1, 4 & 7
or, 3, 4 &5
So, the numbers can be 147, 741, 345 & 543
So, any of the answers are correct.

[saphonymousplayer]

3:53.
147 and 345 should also be included

[iluvsquarez]

-149 is an odd 3 digit number where -1 - 4 = -5 and 4 - 9 = -5 and -1 + 4 + 9 = 12 so that works too

[LeoAr]

For anyone interested, here's another way to solve the limit. You can divide by x on the numerator and the denominator. Then you calculate the limit of each term separately (only possible if each one of those limit exists, which they do). That yields ln(1-x)/x whose limit is -1. You also get sin(x)/x which we know its limit is 1. On the denominator we get sin^2(x)/x. One of the sines with the x gives "1", so we're left with one sine. Finally we arrive to:

lim x->0 -2/sin(x)=+-infinity.

QED

[xnick_uy]

Note that care is required to work with a power series. For instance, the expression ( ln(1-x) + sin(x) ) / (1-cos(x)^2) approaches -1/2 as x approaches 0.

[matheusnb]

Actually, the limit is -infinity or +infinity, depending if it approaches to 0 with a positive nb or an negative

[NoriMori]

I've always wanted to know if these questions made any sense (i.e. whether they actually had correct answers), and whether the answers given were correct. And I'm always playing the sudden death round scene in my head, just because I like it. ("The limit does not exist!") XD Thanks for posting this! Now one of my nagging questions in life has been answered! Edit: Wow. They all make sense _and_ their answers are correct! I have even more respect for the creators now. :D