Question 1152636
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{{{cross(convert 123y=83 base 10)}}} Solve for y: 123 (base y) = 83 base 10.<br>
123 (base y) = y^2+2y+3<br>
Informally, since y is almost certainly a whole number, just do some estimation.  83 is too close to 9^2=81 for y to be 9; so try y=8.<br>
8^2+2(8)+3 = 64+16+3 = 83<br>
ANSWER: y = 8<br>
Formally, we have<br>
{{{y^2+2y+3 = 83}}}
{{{y^2+2y = 80}}}
{{{y^2+2y-80 = 0}}}
{{{(y+10)(y-8) = 0}}}
{{{y = -10}}} or {{{y = 8}}}<br>
While both y=-10 and y=8 are solutions to the equation, we want a positive number for the base, so we reject the solution y = -10.<br>
And although it is overkill for this particular problem, you can find the answer by doing some logical analysis:<br>
(1) y is at least 4, because 1, 2, and 3 are digits in base y.
(2) y is less than 10, because 123 (base 10) is obviously greater then 83.
(3) Since the last digit in base y is 3, 83-3 = 80 must be divisible by y.  That means y is either 4, 5, or 8.<br>
And trying 4 and 5 finds those are too small, so the answer is 8.<br>
That kind of logical reasoning, while not needed for this particular problem, can be useful in similar, more difficult problems.<br>