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The problem is as follows:

In two numbering systems which are of consecutive bases, there are about $120$ three-digit numerals more than the other. Find the smallest of these bases.

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&\textrm{base 11}\\ 2.&\textrm{base 13}\\ 3.&\textrm{base 9}\\ 4.&\textrm{base 15}\\ 5.&\textrm{base 14}\\ \end{array}$

This problem makes me confused. How exactly should it be assessed?. I'm assuming that the author of this problem intends to say that there are two numbers as follows:

$\overline{abc}_{(\alpha)}$ and $\overline{cde}_{(\beta)}$

Then I believe it mentions that there is a difference between $120$ numbers between these digits. Therefore in order to avoid falling into the telephone post error then I think it meant:

$\overline{abc}_{(\alpha)}-\overline{cde}_{(\beta)}=120-1=119$

This assumes that the first number is greater than the second.

But here's the catch, as it stands there are too many variables to guess or poinpoint where to start any sort of inspection which makes the problem to be a guessing game.

Can someone help me with an approapiate method and a step by step approach on how to solve this problem?. It would help me a lot that the answer would be an easy to understand approach.

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1 Answer 1

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We have base $b$ and base $b+1$. In base $b$, there are $b^3-b^2$ three-digit numbers. Likewise, in base $b+1$ there are $(b+1)^3-(b+1)^2$, so this count exceeds that for base $b$ by $$(b+1)^3-(b+1)^2-b^3+b^2= 3b^2+b $$ and this is given to be "about 120". If we ignore the $b$, this means that $3b^2\approx 120$, so $b^2\approx 40 $, and $b$ should be $6$ or $7$. With $b=6$, $3b^2+b=114$, and with $b=7$, $3b^2+b=154$. So it seems that $b=6$ was intended.

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  • $\begingroup$ Howdy! Upon looking in my book the official answer is 13. I don't know how did they got there. What could be wrong?. Perhaps can you explain me?. Is it that there is a typo or maybe something not considered in your solution?. Are we taking into account the fencepost error?. There's an identity you mentioned about the bases $b^{3}-b^{2}$ to be the 3-digit numerals. Can you include a reference for this?. Again, if such is held true, then the rest seems logical but I don't get very well how did they got to the answer?. Can you please attend these questions? $\endgroup$ Commented Nov 18, 2020 at 20:56

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