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Harvey transforms a four-digit number by reversing the order of its digits, subtracting 1 from all digits that are 1 more than a multiple of 3, and adding 1 to all even digits in that order. Harvey obtains 7793 after transformation. Find the number of distinct numbers that he could have transformed. (MAT 2021 Sample Question 2)

I first noted that since all the digits of $7793$ are odd, Harvey must have added $1$ all of the digits, so the number prior to the last transformation was $6682.$ Then, since the second transformation must have turned each digit into a multiple of three, I added $1$ to each $6$ to give me $7782.$ Then, I reversed the digits to get $2877,$ which appears to be the only solution.

However, the sample solutions state that the answer is actually $24.$ Can somebody explain where I messed up?

questionasker
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2 Answers2

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A digit $7$ in the final number may be the result of adding one to the even digit $6$, or by leaving the odd digit $7$ as is.

If we ignore the first step (digit reversal), which is reversible, you may perhaps check what a digit $0$, digit $1$, ... , digit $9$ turn into after the second and third step - which of these become $7$ or $9$ or $3$?

Hagen von Eitzen
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While it states that:

"Harvey obtains $7793$ after transformation"

all we know is that Harvey added $1$ to all the even digits of what he had before. That does not mean "all the digits of that number were even".

Hence he could have obtained, say, $7792$ as a result of his previous operation.

TonyK
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Prime Mover
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  • Not $7792$. Possible intermediate results are $6693$, $6692$, $6683$ and $6682$ – Daniel Mathias Jul 25 '21 at 14:59
  • @DanielMathias After further investigation, maybe. But just from an immediate first glance at the question, and that specific condition, maybe. The point is that OP has misunderstood the basic logic of the question. Perhaps you might like to craft an answer of your own? – Prime Mover Jul 26 '21 at 06:21