Art of Problem Solving
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

1982 AHSME Problems/Problem 26: Difference between revisions

Created page with "== Problem 26 == If the base <math>8</math> representation of a perfect square is <math>ab3c</math>, where <math>a\ne 0</math>, then <math>c</math> equals <math>\text{(A)} ..."
 
Aopspandy (talk | contribs)
editor
151 edits
No edit summary
 
(8 intermediate revisions by 4 users not shown)
Line 8: Line 8:
\text{(D)} 4\qquad  
\text{(D)} 4\qquad  
\text{(E)} \text{not uniquely determined} </math>     
\text{(E)} \text{not uniquely determined} </math>     
== Solution ==
A perfect square will be <math>(8k+r)^2=64k^2+16kr+r^2\equiv r^2\pmod{16}</math> where <math>r=0,1,...,7</math>.
Notice that <math>r^2\equiv 1,4,9,0 \pmod{16}</math>.
Now <math>ab3c</math> in base 8 is <math>a8^3+b8^2+3(8)+c\equiv 8+c\pmod{16}</math>. It being a perfect square means <math>8+c\equiv 1,4,9,0 \pmod{16}</math>. That means that c can only be 1 so the answer is 1 = <math>\boxed{\textbf{(B)}.}</math>.


== Partial and Wrong Solution ==
== Partial and Wrong Solution ==
From the definition of bases we have <math>24+c\equiv k^2 \pmod{64}</math>, where <math>k^2</math> is the perfect square.
From the definition of bases we have <math>k^2=512a+64b+24+c</math>, and <math>k^2\equiv 24+c \pmod{64}</math>


If <math>k=8j</math>, then <math>(8j)^2\equiv64j^2\equiv0 \pmod{64}</math>
If <math>k=8j</math>, then <math>(8j)^2\equiv64j^2\equiv0 \pmod{64}</math>, which makes <math>c\equiv -24\pmod{64}</math>


If <math>k=8j+1</math>, then <math>(8j+1)\equiv 64j^2+16j+1\equiv 16j+1\equiv 24+c \implies 16j\equiv 23+c</math>, which clearly can only have the solution <math>c\equiv 7 \pmod{64}</math>, for <math>j\equiv 1</math>. This makes <math>k=9</math>, which doesn't have 4 digits in base 8
If <math>k=8j+1</math>, then <math>(8j+1)\equiv 64j^2+16j+1\equiv 16j+1\equiv 24+c \implies 16j\equiv 23+c</math>, which clearly can only have the solution <math>c\equiv 7 \pmod{64}</math>, for <math>j\equiv 1</math>. This makes <math>k=9</math>, which doesn't have 4 digits in base 8
Line 18: Line 25:
If <math>k=8j+2</math>, then <math>(8j+1)\equiv 64j^2+32j+4\equiv 32j+4\equiv 24+c \implies 32j\equiv 20+c</math>, which clearly can only have the solution <math>c\equiv 12 \pmod{64}</math>, for <math>j\equiv 1</math>. <math>c</math> is greater than <math>9</math>, and thus, this solution is invalid.  
If <math>k=8j+2</math>, then <math>(8j+1)\equiv 64j^2+32j+4\equiv 32j+4\equiv 24+c \implies 32j\equiv 20+c</math>, which clearly can only have the solution <math>c\equiv 12 \pmod{64}</math>, for <math>j\equiv 1</math>. <math>c</math> is greater than <math>9</math>, and thus, this solution is invalid.  


If <math>k=8j+3</math>, then <math>(8j+1)\equiv 64j^2+48j+9\equiv 48j+9\equiv 24+c \implies 48j\equiv 15+c</math>, which clearly has no solutions for <math>c<10</math>.  
If <math>k=8j+3</math>, then <math>(8j+3)\equiv 64j^2+48j+9\equiv 48j+9\equiv 24+c \implies 48j\equiv 15+c</math>, which clearly has no solutions for <math>0\leq c<10</math>.  
 
Similarly, <math>k=8j+4</math>, yields no solutions


Similarly, <math>k=8j+4</math> yields no solutions


If <math>k=8j+5</math>, then <math>(8j+5)\equiv 64j^2+80j-1\equiv 16j-1\equiv 24+c \implies 16j\equiv 25+c</math>, which clearly can only have the solution <math>c\equiv 9 \pmod{64}</math>, for <math>j\equiv 1</math>. This makes <math>k=13</math>, which doesn't have 4 digits in base 8.  
If <math>k=8j+5</math>, then <math>(8j+5)\equiv 64j^2+80j-1\equiv 16j-1\equiv 24+c \implies 16j\equiv 25+c</math>, which clearly can only have the solution <math>c\equiv 9 \pmod{64}</math>, for <math>j\equiv 1</math>. This makes <math>k=13</math>, which doesn't have 4 digits in base 8.  


If <math>k=8j+6</math>, then <math>(8j+6)\equiv 64j^2+96j+1\equiv 32j+36\equiv 24+c \implies 16j\equiv 23+c</math>, which clearly can only have the solution <math>c\equiv 7 \pmod{64}</math>, for <math>j\equiv 1</math>. This makes <math>k=9</math>, which doesn't have 4 digits in base 8
If <math>k=8j+6</math>, then <math>(8j+6)\equiv 64j^2+96j+1\equiv 32j+36\equiv 24+c \implies 16j\equiv 23+c</math>, which clearly can only have the solution <math>c\equiv 7 \pmod{64}</math>, for <math>j\equiv 1</math>. This makes <math>k=9</math>, which doesn't have 4 digits in base 8
==See Also==
{{AHSME box|year=1982|num-b=25|num-a=27}}
{{MAA Notice}}

Latest revision as of 14:50, 17 June 2021

Problem 26

If the base $8$ representation of a perfect square is $ab3c$, where $a\ne 0$, then $c$ equals

$\text{(A)} 0\qquad \text{(B)}1 \qquad \text{(C)} 3\qquad \text{(D)} 4\qquad \text{(E)} \text{not uniquely determined}$

Solution

A perfect square will be $(8k+r)^2=64k^2+16kr+r^2\equiv r^2\pmod{16}$ where $r=0,1,...,7$.

Notice that $r^2\equiv 1,4,9,0 \pmod{16}$.

Now $ab3c$ in base 8 is $a8^3+b8^2+3(8)+c\equiv 8+c\pmod{16}$. It being a perfect square means $8+c\equiv 1,4,9,0 \pmod{16}$. That means that c can only be 1 so the answer is 1 = $\boxed{\textbf{(B)}.}$.

Partial and Wrong Solution

From the definition of bases we have $k^2=512a+64b+24+c$, and $k^2\equiv 24+c \pmod{64}$

If $k=8j$, then $(8j)^2\equiv64j^2\equiv0 \pmod{64}$, which makes $c\equiv -24\pmod{64}$

If $k=8j+1$, then $(8j+1)\equiv 64j^2+16j+1\equiv 16j+1\equiv 24+c \implies 16j\equiv 23+c$, which clearly can only have the solution $c\equiv 7 \pmod{64}$, for $j\equiv 1$. This makes $k=9$, which doesn't have 4 digits in base 8

If $k=8j+2$, then $(8j+1)\equiv 64j^2+32j+4\equiv 32j+4\equiv 24+c \implies 32j\equiv 20+c$, which clearly can only have the solution $c\equiv 12 \pmod{64}$, for $j\equiv 1$. $c$ is greater than $9$, and thus, this solution is invalid.

If $k=8j+3$, then $(8j+3)\equiv 64j^2+48j+9\equiv 48j+9\equiv 24+c \implies 48j\equiv 15+c$, which clearly has no solutions for $0\leq c<10$.

Similarly, $k=8j+4$ yields no solutions

If $k=8j+5$, then $(8j+5)\equiv 64j^2+80j-1\equiv 16j-1\equiv 24+c \implies 16j\equiv 25+c$, which clearly can only have the solution $c\equiv 9 \pmod{64}$, for $j\equiv 1$. This makes $k=13$, which doesn't have 4 digits in base 8.

If $k=8j+6$, then $(8j+6)\equiv 64j^2+96j+1\equiv 32j+36\equiv 24+c \implies 16j\equiv 23+c$, which clearly can only have the solution $c\equiv 7 \pmod{64}$, for $j\equiv 1$. This makes $k=9$, which doesn't have 4 digits in base 8

See Also

1982 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 25 		Followed by
Problem 27
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions


These problems are copyrighted © by the Mathematical Association of America.

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=1982_AHSME_Problems/Problem_26&oldid=156247"