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Difference between revisions of "2019 AIME II Problems/Problem 3"

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==Problem==
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Find the number of <math>7</math>-tuples of positive integers <math>(a,b,c,d,e,f,g)</math> that satisfy the following systems of equations:
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<cmath>\begin{align*}
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abc&=70,\\
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cde&=71,\\
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efg&=72.
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\end{align*}</cmath>
  
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==Solution 1==
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As 71 is prime, <math>c</math>, <math>d</math>, and <math>e</math> must be 1, 1, and 71 (in some order). However, since <math>c</math> and <math>e</math> are divisors of 70 and 72 respectively, the only possibility is <math>(c,d,e) = (1,71,1)</math>. Now we are left with finding the number of solutions <math>(a,b,f,g)</math> satisfying <math>ab = 70</math> and <math>fg = 72</math>, which separates easily into two subproblems. The number of positive integer solutions to <math>ab = 70</math> simply equals the number of divisors of 70 (as we can choose a divisor for <math>a</math>, which uniquely determines <math>b</math>). As <math>70 = 2^1 \cdot 5^1 \cdot 7^1</math>, we have <math>d(70) = (1+1)(1+1)(1+1) = 8</math> solutions. Similarly, <math>72 = 2^3 \cdot 3^2</math>, so <math>d(72) = 4 \times 3 = 12</math>.
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Then the answer is simply <math>8 \times 12 = \boxed{096}</math>.
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-scrabbler94
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==Solution 2==
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We know that any two consecutive numbers are coprime. Using this, we can figure out that <math>c=1</math> and <math>e=1</math>. <math>d</math> then has to be <math>71</math>. Now we have two equations left. <math>ab=70</math> and <math>fg=72</math>. To solve these we just need to figure out all of the factors. Doing the prime factorization of <math>70</math> and <math>72</math>, we find that they have <math>8</math> and <math>12</math> factors, respectively. The answer is <math>8 \times 12=\boxed{096}</math>
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~kempwood
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==See Also==
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{{AIME box|year=2019|n=II|num-b=2|num-a=4}}
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{{MAA Notice}}

Latest revision as of 20:05, 23 December 2023

Problem

Find the number of $7$-tuples of positive integers $(a,b,c,d,e,f,g)$ that satisfy the following systems of equations: \begin{align*} abc&=70,\\ cde&=71,\\ efg&=72. \end{align*}

Solution 1

As 71 is prime, $c$, $d$, and $e$ must be 1, 1, and 71 (in some order). However, since $c$ and $e$ are divisors of 70 and 72 respectively, the only possibility is $(c,d,e) = (1,71,1)$. Now we are left with finding the number of solutions $(a,b,f,g)$ satisfying $ab = 70$ and $fg = 72$, which separates easily into two subproblems. The number of positive integer solutions to $ab = 70$ simply equals the number of divisors of 70 (as we can choose a divisor for $a$, which uniquely determines $b$). As $70 = 2^1 \cdot 5^1 \cdot 7^1$, we have $d(70) = (1+1)(1+1)(1+1) = 8$ solutions. Similarly, $72 = 2^3 \cdot 3^2$, so $d(72) = 4 \times 3 = 12$.

Then the answer is simply $8 \times 12 = \boxed{096}$.

-scrabbler94

Solution 2

We know that any two consecutive numbers are coprime. Using this, we can figure out that $c=1$ and $e=1$. $d$ then has to be $71$. Now we have two equations left. $ab=70$ and $fg=72$. To solve these we just need to figure out all of the factors. Doing the prime factorization of $70$ and $72$, we find that they have $8$ and $12$ factors, respectively. The answer is $8 \times 12=\boxed{096}$

~kempwood

See Also

2019 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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