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Difference between revisions of "1968 AHSME Problems/Problem 33"

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== Solution ==
 
== Solution ==
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Call the number abc.
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Call the number ''abc''.
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Then, 49a+7b+c=81c+9b+a. (Breaking down the number in base-form)
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Then, 49''a''+7''b''+''c''=81''c''+9''b''+''a''. (Breaking down the number in base-form);
−
Moving everything to one side,  
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After combining like terms and moving the variables around,
−
48a=2b+80c, b=40c-24a=8(5c-2a). This shows that b is a multiple of 8. Thus, b=0 (since 8>7, the base of base 7).
+
48''a''=2''b''+80''c'', ''b''=40''c''-24''a''=8(5''c''-2''a''). This shows that ''b'' is a multiple of 8 (we only have to find the middle digit under ''one'' of the bases). Thus, ''b''=0 (since 8>6, the largest digit in base 7).
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So b=0. Select "A."
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So ''b''=0.
−
I have no idea how to format, so please help. Thanks.
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Select <math>\fbox{A}</math> as our answer.
−
 
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I have no idea how to format besides the basics, so please help. Feel free to message me tips on AOPS. Thanks.
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<math>\fbox{A}</math>
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~hastapasta
  
 
== See also ==
 
== See also ==

Revision as of 17:09, 4 January 2022

Problem

A number $N$ has three digits when expressed in base $7$. When $N$ is expressed in base $9$ the digits are reversed. Then the middle digit is:

$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) } 5$

Solution

Call the number abc. Then, 49a+7b+c=81c+9b+a. (Breaking down the number in base-form); After combining like terms and moving the variables around, 48a=2b+80c, b=40c-24a=8(5c-2a). This shows that b is a multiple of 8 (we only have to find the middle digit under one of the bases). Thus, b=0 (since 8>6, the largest digit in base 7). So b=0. Select $\fbox{A}$ as our answer. I have no idea how to format besides the basics, so please help. Feel free to message me tips on AOPS. Thanks. ~hastapasta

See also

1968 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 32 		Followed by
Problem 34
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. AMC logo.png

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