<PRE><B>Question 1195856</B>
&lt;br&gt;
The problem is this:&lt;br&gt;
321 (base n) = 232 (base 7)&lt;br&gt;
Informally....&lt;br&gt;
In any base (where the numbers are defined), 321 is greater than 232; so if 321 (base n) is equal to 232 (base 7), then the base n must be less than 7.&lt;br&gt;
So try n=6....&lt;br&gt;
321 (base 6)
3*6=18; 18+2=20
20*6=120; 120+1=121&lt;br&gt;
232 (base 7)
2*7=14; 14+3=17
17*7=119; 119+2=121&lt;br&gt;
n=6 works!&lt;br&gt;
ANSWER: n=6&lt;br&gt;
Formally....&lt;br&gt;
321 (base n) = 3n^2+2n+1
232 (base 7) = 2(7^2)+3(7)+2 = 98+21+2 = 121&lt;br&gt;
3n^2+2n+1=121
3n^2+2n-120 = 0
(n-6)(3n+20) = 0&lt;br&gt;
n=6 or n=-20/3&lt;br&gt;
Obviously the negative fractional answer makes not sense.  So&lt;br&gt;
ANSWER: n=6&lt;br&gt;
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