The Correct Guess

My dog takes the AIME, and sees a problem. The question's answer was m n \frac{m}{n} and asked for m + n m+n , where m m and n n are relatively prime positive integers. She randomly guesses an m m and an n n such that they are relatively prime positive integers, and adds them up, and gets the answer 504 504 to be correct. Knowing she got it correct, what is the probability that she guessed the correct m m and the correct n n ?

1 144 \frac 1{144} 1 42 \frac 1{42} 1 24 \frac 1{24} 1 504 \frac 1{504} 1 252 \frac 1{252}

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2 solutions

Similar explanation as @Arindam Kulkarni's, but with more details.

As mentioned by Arindam, the problem is effectively asking how many two relatively prime positive integers or coprime positive integers add up to 504 504 . The integers m m and n n are coprime integers if their greatest common divisor is 1 1 or gcd ( m , n ) = 1 \gcd(m,n)=1 . That is m m and n n don't share any common divisor other than 1 1 . To find the number of positive integers that are coprime to 504 504 , we use the Euler's totient function ϕ ( 504 ) \phi (504) which is computed as follows:

ϕ ( n ) = n ( 1 1 p 1 ) ( 1 1 p 2 ) ( 1 1 p 3 ) ( 1 1 p k ) where p 1 , p 2 , p 3 , p k are all the prime divisors of n . ϕ ( 504 ) = 504 × 1 2 × 2 3 × 6 7 = 144 Since 504 = 2 3 × 3 3 × 7 \begin{aligned} \phi (n) & = n \left(1-\frac 1{p_1}\right)\left(1-\frac 1{p_2}\right)\left(1-\frac 1{p_3}\right)\cdots \left(1-\frac 1{p_k}\right) & \small \blue{\text{where }p_1, p_2, p_3, \cdots p_k \text{ are all the prime divisors of }n.} \\ \implies \phi(504) & = 504 \times \frac 12 \times \frac 23 \times \frac 67 = 144 & \small \blue{\text{Since }504 = 2^3 \times 3^3 \times 7} \end{aligned}

Since Arindam's dog makes 1 1 guess out of 144 144 possibilities, the probability of guessing the right answer is 1 144 \boxed{\frac 1{144}} .

I suppose, you need to divide 144 144 by the total number of possibilities, which is 504 504 . Please let me know if Im wrong.

Alexander Shannon - 1 year, 2 months ago

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The total number of ways is 144 144 and there is only 1 guess. Therefore, the probability is 1 144 \frac 1{144} .

Chew-Seong Cheong - 1 year, 2 months ago

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it is clear now. Thanks!

Alexander Shannon - 1 year, 2 months ago

No. This is a good point. However, notice how the problem says that the dog chooses m and n such that they are already relatively prime and positive integers. Those add to 504. Therefore, we are only looking at those 144 valid cases. There is only 1 case which would have the correct m and n, because that's how problems work, and therefore that is 1/144.

Arindam Kulkarni - 1 year, 2 months ago

I think you may be double-counting. {5,499} and {499,5} aren't distinct solutions.

Richard Desper - 1 year, 2 months ago

The answer of the problem is of the form m n \dfrac mn and 5 499 499 5 \dfrac 5{499} \ne \dfrac {499}5 . The dog is guessing one m m and one n n .

Chew-Seong Cheong - 1 year, 2 months ago

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No, the question says "What is the probability that the dog guessed 'the m and n' correct"? Implying that the pair is correct. As opposed to "What is the probability that the dog guessed m and n correct?"

At least, that's how I read it. The phrasing implies we are selecting from a a set of pairs. Not a set of ordered pairs.

Richard Desper - 1 year, 2 months ago

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I read otherwise.

Chew-Seong Cheong - 1 year, 2 months ago

If m is 499, and n is 5, the answer would be 499/5. However, if m is 5, and n is 499, the answer is 5/499. Therefore, guessing them vice versa is NOT guessing the correct m AND the correct n.

Arindam Kulkarni - 1 year, 2 months ago

Yes, as Chew-Seong Cheong said, m/n for as 499/5 and 5/499 is completely different. Therefore, there is no double counting involved.

Arindam Kulkarni - 1 year, 2 months ago

If m is 499, and n is 5, the answer would be 499/5. However, if m is 5, and n is 499, the answer is 5/499. Therefore, guessing them vice versa is NOT guessing the correct m AND the correct n.

Arindam Kulkarni - 1 year, 2 months ago

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It's an issue of grouping things in language. When I see "the correct m and n" I don't think that means the same as "the correct m and the correct n". I see that as "the correct pair: {m,n}."

But don't worry too much about it - apparently nobody else is reading it this way.

Richard Desper - 1 year, 2 months ago

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Ok. I see your point. I will change it for good measure.

Arindam Kulkarni - 1 year, 2 months ago

Hi, sorry i'm reviewing for this topic, but i cant seem to understand why the solution above looks for integers coprime to 504 if we're looking for integers that add up to 504? Am i missing something in my train of thought?

Harvey Felipe - 1 year, 2 months ago

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Yeah. So imagine on an AIME, you get an answer like 3/18, and it asks for fully simplified form, and add the numerator and denominator. If you put 3+18 = 21, you get it WRONG as it isn't simplified (I almost did that in this year's AIME). The correct answer would be 1/6 and 1+6 = 7. Therefore, not all pairs that add up to 504 are eligible as my dog's m and n, as the problem states m and n are both chosen correctly.

Now, let's see when a case is INVALID. say m and n are ak and bk, where k is a multiple of a prime. in that case, 504 = k(a+b). As a result, if m and n are NOT In simplest form, they have a prime factor in common with 504! This is why we find relatively prime integers to 504. How we do that is shown in the other solutions.

Arindam Kulkarni - 1 year, 2 months ago
Arindam Kulkarni
Apr 2, 2020

The problem is effectively asking how many ways can two relatively prime positive integers add up to 504. If m and n are NOT relatively prime, their GCD can be any factor of 504. This means you want to find how many numbers are relatively prime to 504. This is simply φ(504). This is 504 * 6 7 \frac{6}{7} * 1 2 \frac{1}{2} * 2 3 \frac{2}{3} = 144 .

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