Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2017 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 2"

(→‎Solution)
(→‎Solution)
Line 8: Line 8:
 
<math>A = 3, T = 9, S = 1</math> and <math>R = 2</math>
 
<math>A = 3, T = 9, S = 1</math> and <math>R = 2</math>
  
−
The question states that
+
Another way of stating the question is that <math>SAT</math> <math>\cdot</math> <math>SAT</math> = <math>STARS</math>.
 +
This means that the ending digit of <math>T</math> <math>\cdot</math> <math>T</math> or <math>S</math>  has to be the ending digit of a square number, or {<math>0, 1, 4, 5, 6, 9</math>}. Keep in mind that <math>STARS</math> is a 5-digit number. We know that <math>S</math> cannot be 0 because <math>SAT</math> would become a 2-digit number. It cannot be 4 or over because <math>STARS</math> would have 6 digits instead of 5. This only leaves <math>S</math> = 1. Because the last digit of <math>STARS</math> is <math>1</math>, we can conclude that <math>T</math> must be <math>9</math> or <math>1</math>. However, since all the digits are unique, <math>T</math> = <math>9</math>.
 +
Now the expression becomes:
 +
 
 +
<math>1A9</math> <math>\cdot</math> <math>1A9</math> = <math>19AR1</math>
 +
 
 +
Because the first two digits of <math>STARS</math> are <math>1</math> and <math>9</math>, A cannot take on too many value because if it becomes 5, <math>S</math> will become 2. It cannot be four either because even then, <math>STARS</math> would be too small. Now, if A is tested from the set {<math>0,1,2,3</math>}:
 +
 
 +
<math>A=0</math>:
 +
 
 +
<math>109</math> <math>\cdot</math> <math>109</math> = <math>11881</math>  (Does not work)
 +
 
 +
<math>A=1</math>:
 +
 
 +
<math>119</math> <math>\cdot</math> <math>119</math> = <math>14161</math>  (Does not work)
 +
 
 +
<math>A=2</math>:
 +
 
 +
<math>129</math> <math>\cdot</math> <math>129</math> = <math>16641</math>  (Does not work)
 +
 
 +
<math>A=3</math>:
 +
 
 +
<math>139</math> <math>\cdot</math> <math>139</math> = <math>19321</math>
 +
 
 +
This works as it fits the template: <math>\boxed{\textbf{(S=1, T=9, A=3, and R=2)}}</math>
  
 
== See also ==
 
== See also ==

Revision as of 12:22, 21 August 2023


Problem

Suppose $A, R, S$, and $T$ all denote distinct digits from $1$ to $9$. If $\sqrt{STARS} = SAT$, what are $A, R, S$, and $T$?

Solution

$A = 3, T = 9, S = 1$ and $R = 2$

Another way of stating the question is that $SAT$ $\cdot$ $SAT$ = $STARS$. This means that the ending digit of $T$ $\cdot$ $T$ or $S$ has to be the ending digit of a square number, or {$0, 1, 4, 5, 6, 9$}. Keep in mind that $STARS$ is a 5-digit number. We know that $S$ cannot be 0 because $SAT$ would become a 2-digit number. It cannot be 4 or over because $STARS$ would have 6 digits instead of 5. This only leaves $S$ = 1. Because the last digit of $STARS$ is $1$, we can conclude that $T$ must be $9$ or $1$. However, since all the digits are unique, $T$ = $9$. Now the expression becomes:

$1A9$ $\cdot$ $1A9$ = $19AR1$

Because the first two digits of $STARS$ are $1$ and $9$, A cannot take on too many value because if it becomes 5, $S$ will become 2. It cannot be four either because even then, $STARS$ would be too small. Now, if A is tested from the set {$0,1,2,3$}:

$A=0$:

$109$ $\cdot$ $109$ = $11881$ (Does not work)

$A=1$:

$119$ $\cdot$ $119$ = $14161$ (Does not work)

$A=2$:

$129$ $\cdot$ $129$ = $16641$ (Does not work)

$A=3$:

$139$ $\cdot$ $139$ = $19321$

This works as it fits the template: $\boxed{\textbf{(S=1, T=9, A=3, and R=2)}}$

See also

2017 UNM-PNM Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNM-PNM Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_UNM-PNM_Statewide_High_School_Mathematics_Contest_II_Problems/Problem_2&oldid=197271"