document.write( "Question 1152636: convert 123y=83 base 10
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Algebra.Com's Answer #774747 by greenestamps(13200) $\"\"$ $\"About$

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\n" ); document.write( " $\"cross%28convert+123y=83+base+10%29\"$ Solve for y: 123 (base y) = 83 base 10.

\n" ); document.write( "123 (base y) = y^2+2y+3

\n" ); document.write( "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.

\n" ); document.write( "8^2+2(8)+3 = 64+16+3 = 83

\n" ); document.write( "ANSWER: y = 8

\n" ); document.write( "Formally, we have

\n" ); document.write( " $\"y%5E2%2B2y%2B3+=+83\"$
\n" ); document.write( " $\"y%5E2%2B2y+=+80\"$
\n" ); document.write( " $\"y%5E2%2B2y-80+=+0\"$
\n" ); document.write( " $\"%28y%2B10%29%28y-8%29+=+0\"$
\n" ); document.write( " $\"y+=+-10\"$ or $\"y+=+8\"$

\n" ); document.write( "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.

\n" ); document.write( "And although it is overkill for this particular problem, you can find the answer by doing some logical analysis:

\n" ); document.write( "(1) y is at least 4, because 1, 2, and 3 are digits in base y.
\n" ); document.write( "(2) y is less than 10, because 123 (base 10) is obviously greater then 83.
\n" ); document.write( "(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.

\n" ); document.write( "And trying 4 and 5 finds those are too small, so the answer is 8.

\n" ); document.write( "That kind of logical reasoning, while not needed for this particular problem, can be useful in similar, more difficult problems.

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