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

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== Solution==
 
== Solution==
 
<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>
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Another way of stating the question is that <math>SAT</math> <math>\cdot</math> <math>SAT</math> = <math>STARS</math>.
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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>.
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Now the expression becomes:
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<math>1A9</math> <math>\cdot</math> <math>1A9</math> = <math>19AR1</math>
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Because the first two digits of <math>STARS</math> are <math>1</math> and <math>9</math>, <math>A</math> cannot take on too many values because if it becomes <math>5</math>, <math>S</math> will become <math>2</math>. It cannot be <math>4</math> either because even then, <math>STARS</math> would be too small to satisfy (If <math>SAT=140</math>, it would still have too big a square for <math>STARS</math> to be able to satisfy, <math>A</math> being <math>4</math>). Now, if <math>A</math> is tested from the set {<math>0,1,2,3</math>}:
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<math>A=0</math>:
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<math>109</math> <math>\cdot</math> <math>109</math> = <math>11881</math>  (Does not work)
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<math>A=1</math>:
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<math>119</math> <math>\cdot</math> <math>119</math> = <math>14161</math>  (Does not work)
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<math>A=2</math>:
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<math>129</math> <math>\cdot</math> <math>129</math> = <math>16641</math>  (Does not work)
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<math>A=3</math>:
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<math>139</math> <math>\cdot</math> <math>139</math> = <math>19321</math>
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This works as it fits the template: <math>\boxed{\textbf{(S=1, T=9, A=3, and R=2)}}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 12:27, 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 values because if it becomes $5$, $S$ will become $2$. It cannot be $4$ either because even then, $STARS$ would be too small to satisfy (If $SAT=140$, it would still have too big a square for $STARS$ to be able to satisfy, $A$ being $4$). 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
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