Difference between revisions of "2017 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 2"
(Created page with " == Problem == Suppose <math>A, R, S</math>, and <math>T</math> all denote distinct digits from <math>1</math> to <math>9</math>. If <math>\sqrt{STARS} = SAT</math>, what ar...") |
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== Solution== | == Solution== | ||
+ | <math>A = 3, T = 9, S = 1</math> and <math>R = 2</math> | ||
+ | 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>, <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>}: | ||
+ | |||
+ | <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 == |
Latest revision as of 12:27, 21 August 2023
Problem
Suppose , and all denote distinct digits from to . If , what are , and ?
Solution
and
Another way of stating the question is that = . This means that the ending digit of or has to be the ending digit of a square number, or {}. Keep in mind that is a 5-digit number. We know that cannot be 0 because would become a 2-digit number. It cannot be 4 or over because would have 6 digits instead of 5. This only leaves = 1. Because the last digit of is , we can conclude that must be or . However, since all the digits are unique, = . Now the expression becomes:
=
Because the first two digits of are and , cannot take on too many values because if it becomes , will become . It cannot be either because even then, would be too small to satisfy (If , it would still have too big a square for to be able to satisfy, being ). Now, if is tested from the set {}:
:
= (Does not work)
:
= (Does not work)
:
= (Does not work)
:
=
This works as it fits the template:
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 |