There are 2 faulty odometer problems posted.

I believe that these problems can be solved by taking the numbers shown on the odometers as numbers in the base 9 and finding their values in base 10. For example: In the problem where the odometer skips 9 the faulty odometer registers 120. Changing this to base 10 yields 99 which was given as the answer.

In the problem where the odometer skips 8 the number registered was 2197. If we change this to base 10 we get 1627 which is the answer I got by a different method although it doesn't match the alleged answer of 1728.

Any thoughts?

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## Comments

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TopNewestGeneral theory: In the "Faulty Odometer" type of problem, if the excluded digit is "d" then every digit greater than "d" in the faulty reading should be decreased by 1. This number taken as written in base 9 and expressed in base 10 will be the true reading.

For example: If the excluded digit is 3 and the faulty reading is 1428 then this should be changed to 1327. Expressed in base 10 we get 997.

If the excluded digit is 1 and the faulty reading is 35 then this should be changed to 24. Expressed in base 10 the true reading is 22.

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Yup, this is the approach to take. Can you give a quick explanation of why this works?

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Which question are you telling?????

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Just search "Faulty Odometer". There are 2 problems with the same names.

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My latest answer by a different method is 1618 which negates my theory.

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Your claim is that: "If the odometer skips the number X, and shows the number Y in base 10, then the actual distance moved is Y read in base 9. This is independent of the number X".

Why is this claim not true?

For what fixed value of X, if any, is the claim true?

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Sorry, I don't understand your comment.

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My comment is that: when all we did was "change the base to 9 and replace any substituted digit by the excluded digit" doesn't work for all "X".

In particular, it does not work for \( X = 9 \), which used the algorithm that you originally stated.

This algorithm (of replacing, instead of decreasing every relevant digit), only works for \( X = 8 \).

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To summarize,

Algorithm 1 (given in note) : Calculate in Base 9.

This only works for \(X=9\).

Algorithm 2 (given in comment on July 13): Replace the digits 9 with the substituted digit X, and calculate in Base 9.

This only works for \( X = 8 \). [Technically it also works for \(X=9\), since there is no replacement happening]

Algorithm 3 (given in comment 21 hours ago): Replace the digits \( X+i\) with \( X + i-1 \) for all positive \(i\), and calculate in Base 9.

This works for all \(X\).

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Further thoughts (revelations?): My theory works provided I replace any substituted digit by the excluded digit in the number to the base 9 and expressing the number to base 10 will get the answer that was found by using the other method.

For example: skipping the digit 8: 121=100, 242=200, 363=300, 494=400. The first 3 answers can be obtained by my short-cut method however the last does not. It will however work if 494 is replaced by 484.

The number 2197 changed to 2187 and converted to base 10 gives 1618, the answer I obtained by the other method.

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What if the odometer will not register an odd digit?

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