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After applying a tax of 13%, certain values are impossible. Find those below $5.00$

For example$0.946$ would round to $0.95$

I know how to use a brute force approach but there has to be a better way. I know applying the tax is just multiplying our number by $1.13$ but I don’t get it how to do it faster other than use brute force.

We are learning vectors so it might have to do something with that. Also $5.00$ is impossible (I checked).

Edit: the question basically says real numbers of the form $x.yz$ (A number followed by two decimal places)when multiplied by $1.13$ do not include certain numbers less than $5.00$ when we round up the hundredths. For instance, $1.145$ rounds to $1.15$.

Dhdh
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  • What do you mean by "impossible"? Please define the mathematical equation that needs to be solved. I think what you're after is something with rounding. Are you looking for numbers that round to 5.00, for example? At what level are we rounding? Tenths? Hundredths? – Matti P. Feb 14 '20 at 13:34
  • @MattiP.: It's pretty clear what is meant; certain amounts of dollars and cents cannot be reached after applying a tax of 13% and rounding. For instance, 80¢ * 113% = 90.4 rounds to 90¢, and 81¢ * 113% = 91.53 rounds to 92¢, so it is impossible to get 91¢ after a 13% tax and rounding. – Nick Matteo Mar 29 '20 at 00:12

1 Answers1

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Let's just look at cents, so that we can deal with whole numbers. So the question is which numbers between 0 and 500, exclusive, are not the result of rounding $n*1.13$ for any natural number $n$.

Since the 100 numbers from 1 to 100 are mapped to 113 numbers from 1 to 113, we know that 13 numbers are left out. After that, 101 * 1.13 = 113 + 1*1.13, so the images repeat, modulo 113.

Thus, if we determine that the first 13 omitted numbers are 4, 13, 22, 30, 39, 48, 57, 65, 74, 83, 91, 100, and 109, we know the next 13 are 113 + 4, 113 + 13, etc., and the following 13 are 226 + 4, etc., ending with $452 + 39 = 491$, a total of $13\cdot4+5 = 57$ omitted values.

Another way to think about it: We skip a number $k+1$ if $1.13n < k + 0.5$ and $1.13(n+1) \geq k + 1.5$. Scaling up to whole numbers, the latter condition is $113n + 113 \geq 100k + 150$, or $113n \geq 100k + 37$.

Combining the conditions, we have $100k + 37 \leq 113n < 100k + 50$. So $13n$ is between 37 and 49 modulo 100. It turns out 77 is the multiplicative inverse of 13 mod 100, since 13*77 = 1001. So the values of $n$ with $13n \equiv 37, \dotsc, 49 \pmod{100}$ are $$37\cdot77,38\cdot77,\dotsc,49\cdot77 \pmod {100}$$ which have least residues $$49, 26, 3, 80, 57, 34, 11, 88, 65, 42, 19, 96, 73.$$ So these 13 values, and then 100 plus each, 200 plus each, etc., are the preimages whose images precede a skipped value.

Nick Matteo
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