Difference between revisions of "1982 AHSME Problems/Problem 26"
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\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>. | 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>. | 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>. | + | 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 == | ||
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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 15:50, 17 June 2021
Problem 26
If the base representation of a perfect square is
, where
, then
equals
Solution
A perfect square will be where
.
Notice that .
Now in base 8 is
. It being a perfect square means
. That means that c can only be 1 so the answer is 1 =
.
Partial and Wrong Solution
From the definition of bases we have , and
If , then
, which makes
If , then
, which clearly can only have the solution
, for
. This makes
, which doesn't have 4 digits in base 8
If , then
, which clearly can only have the solution
, for
.
is greater than
, and thus, this solution is invalid.
If , then
, which clearly has no solutions for
.
Similarly, yields no solutions
If , then
, which clearly can only have the solution
, for
. This makes
, which doesn't have 4 digits in base 8.
If , then
, which clearly can only have the solution
, for
. This makes
, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.