Difference between revisions of "2019 AIME II Problems/Problem 3"
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\end{align*}</cmath> | \end{align*}</cmath> | ||
| − | ==Solution== | + | ==Solution 1== |
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>. | 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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-scrabbler94 | -scrabbler94 | ||
| + | ==Solution 2== | ||
| + | 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 71. 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{96}</math> | ||
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| + | ~Hithere22702 | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=II|num-b=2|num-a=4}} | {{AIME box|year=2019|n=II|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:17, 28 April 2020
Contents
Problem 3
Find the number of 
-tuples of positive integers 
that satisfy the following systems of equations:
Solution 1
As 71 is prime, 
, 
, and 
must be 1, 1, and 71 (in some order). However, since and are divisors of 70 and 72 respectively, the only possibility is 
. Now we are left with finding the number of solutions 
satisfying 
and 
, which separates easily into two subproblems. The number of positive integer solutions to simply equals the number of divisors of 70 (as we can choose a divisor for 
, which uniquely determines 
). As 
, we have 
solutions. Similarly, 
, so 
.
Then the answer is simply 
.
-scrabbler94
Solution 2
We know that any two consecutive numbers are coprime. Using this, we can figure out that 
and 
. then has to be 71. Now we have two equations left. 
and 
. To solve these we just need to figure out all of the factors. Doing the prime factorization of 
and 
, we find that they have 
and 
factors, respectively. The answer is 
~Hithere22702
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
