# A Change in Code

All numbers in the first column have been increased by the same percentage, to give the results shown in the second column.

*A given letter stands for the same numeral every time it appears in that column, so for example, whatever the C stands for, it stands for that numeral every time the C appears.*

**Can you work out the percentage increase?**

Notice that all the numbers in the first column are multiples of 5, and all the numbers in the second column are whole numbers.

Notice that when 80 is increased by the secret percentage, the result is a two-digit number.

**What about this next set?**

It's a new coding - the letters now stand for different numerals.

What percentage have the numbers in the first column been increased by this time?

Notice that all the numbers in the second column are multiples of 5.

Notice that when the numbers in the first column are increased by the secret percentage, the result is almost always a two-digit number.

**Can you work out the percentage increase here?**

Different coding again, and much harder...

Do the numbers in the second column have a common factor?

Notice that whole numbers have stayed as whole numbers after the increase.

What could all the original numbers be divided by without producing a decimal anywhere in the results column?

**The first table**

Emily and Kaia from Stoke-by-Nayland Middle School tried out some numbers for the percentage change:

We started by looking at it for a obvious pattern. One thing that was obvious was that there were no decimal places,they were all whole numbers and two-digits. So we thought we'd try 10% with 15, but it turned out as a decimal-16.5. So it was an obvious step to do 20% next. It turned out to be a whole number. We then realised that 20's code was 35 turned around. We worked out 20's code and it was a whole number and then worked out 35's code and it was 20's code switched around. We then worked out all the number at 20% and found that they all worked.

Hari chose a couple of the examples from the table and used those to check for possible increases:

$x\div100\times20=ab$

$x\div100\times35=ba$

I did trial and error

$120\div100\times20=24$

$120\div100\times35=42$

Paul from Castleknock College, Dublin, Jeremy from Drexel Univesity and Ben first found the smallest and largest values that the percentage increase could take. This is Ben's work:

To work out the first percentage, it is possible to begin by working out bounds for the possible solution. All the final numbers are 2-digit numbers, and the largest starting number is 80. This cannot be increased to a number equal to or greater than 100 or it would cease to be a 2-digit number. 100/80 = 1.25 so the percentage increase must be less than 25%.

Now consider a minimum percentage increase by finding two starting numbers close together: say 20 and 25. These are turned into GC and EB; note that the first digit in the final numbers must thus be different. So if we multiply them both by a small percentage change, for example 10% (1.1) then the final numbers are 20 $\times$ 1.1 = 22 and 25 $\times$ 1.1 = 27.5 (ignore for a minute the fact that the final numbers must be integers). But 25 and 27.5 both have a 2 as the first digit - and the coded answers show they must begin with different digits. In fact, for them to begin with different digits the minimum percentage change is 20%, as 25 $\times$ 1.2 = 30, and 20 $\times$ 1.2 = 24, 20 being just high enough for the E to represent a greater digit than the G. So 20% is the minimum percentage increase and 25% is the maximum percentage increase.

To find the exact value, consider that that each starting number multiplied by the percentage change must equal an integer. If we look at 15 as the starting number, increasing this by 20% gives 15 $\times$ 1.2 = 18, so 20% is a possibility. Are there any other possibilities? The next integer IJ could be is 19, but what percentage increase would this require? To find this, we divide 19 by 15, which gives 1.267, or a 26.7% percentage increase. This is greater than 25%, the maximum value, so the only possible solution is a 20% increase! Multiplying through all the numbers by 1.2 allows the code to be deduced:

B=0, I=1, G=2, E=3, C=4, A=5, F=6, D=7, J=8 and H=9.

Paul's reasoning was very similar, but Paul thought about common factors (and had a tighter condition on the largest possible percentage increase):

The answer to the first part is a 20% increase. The most it can be is a 22.5% increase because 80 would then become 98, it can't 99 or else there would be 2 equal letters. All numbers have one thing in common they're divisible by 5 so therefore the only way all of these numbers could become perfect integers is if they were multiplied by $\frac15.$ It must be 20% because anything greater, 40% 60% etc., would turn 80 into a three digit number which it clearly isn't.

Jeremy also found the highest and lowest values for the percentage increase, but also considered the possibility that the percentage increase could actually be a decrease. Then, Jeremy used results with the same last digits to work out what the increase was:

Call the given positive integers $a_0,a_2,...,a_7$

Call the percentage asked for as a real number $p$

Call the result as a positive integer $r_0,r_2,...,r_7$ where $r_n = a_n \times p$ c

Call the coded results $c_0,c_1,...,c_7$

1) Determine the possible range of $p$

$r_n$ must be 2 digits, so you know each $r_n$ will be $10 \le r_n \lt 100.$

Because $r_n = a_n \times p, 10 \le a_n \times p \lt 100,$ and therefore:

$10 \div \min{(a_n)}\le p \lt 100 \div \max{(a_n)},$ since each $a_n$ is always positive.

$\min{(a_n)}=a_1=15$ and $\max{(a_n)}=a_5=80.$ Therefore:

$10 \div 15 \le p \lt 100 \div 80$

$0.66 \le p \lt 1.25$

2) Examine the coded results and relate digits in results showing a pattern to each other (note: a digit in a result can be related to another digit in the same result).

The examples for this step only show one relation per problem and are for simplicity. The actual deduction of p gets easier the more relations that you make.

The second digit of $c_0, c_2,$ and $c_4$ is 'C'.

$(r_0 \mod 10)=(r_2 \mod 10)=(r_4 \mod 10)$ (*or, in other words, the remainder when $r_0$ is divided by $10$ is equal to the remainder then $r_2$ is divided by $10$ and to the remainder then $r_4$ is divided by $10$*)

$(a_0\times p \mod 10)=(a_2\times p \mod 10)=(a_4\times p \mod 10)$

$(20\times p \mod 10)=(45\times p \mod 10)=(70\times p \mod 10)$

3) Find a p (within range) that satisfies the relation and produces integers.

$p=1.2$ (*since $0.66 \le 1.2 \le 1.25,$ $(20\times 1.2 \mod 10)=(45\times 1.2 \mod 10)=(70\times 1.2 mod 10)=4$ (because the last digit in each case is $4,$ and this is also the remainder when they are divided by $10$)*)

**The second table**

Ben used a similar method to before to calculate the lower limit for the percentage change, and then used trial from there:

Again, in the 2nd part lets establish a minimum percentage change. All the starting numbers have 2 digits, and the largest end number is 120, so this must decrease to under 100. What's more it can't be 99 as the 2 digits are different, so the maximum AJ can be is 98. So the minimum percentage change is an increase of 120$\div$98 = 22.4%.

Now we need to consider a maximum increase, so lets find two end numbers close together - 45 and 50 look promising. The start numbers are IJ and ED, again note that the first digits are different. Using a similar technique to before, lets consider the possible percentage changes. Start by dividing them both by 1.224, the minimum increase, giving 36.8 for 45 and 40.8 for 50. So lets try E = 4
and I = 3 for now. If E = 4, D must be 0 as if ED was 41, the percentage change would be 50$\div$41 = 1.220, and this is less than our minimum. If ED is 40, the percentage change is 50$\div$40 = 1.25. Note that at this point we do not know that E is 4 and I is 3, E could be 3 and I could be 2, but it would take some time to examine all these possibilities, so lets try dividing all the end
numbers by 1.25 and see if this seems to fit:

65$\div$1.25 = 52

35$\div$1.25 = 28

80$\div$1.25 = 64

120$\div$1.25 = 96

15$\div$1.25 = 12

45$\div$1.25 = 36

50$\div$1.25 = 40

95$\div$1.25 = 76

This provides integer solutions that fit the code D=0, B=1, C=2, I=3, E=4, F=5, J=6, H=7, G=8 and A=9, so the percentage changed used on the first column numbers is an increase of 25%.

Jeremy found the lower limit in the same way as Ben, and also found the upper limit. Jeremy then continued using the same method and notation as for the first table:

For problems 2 and 3, the naming conventions are the same but since we are going backwards, $r_n = a_n \div p.$

1) Determine the possible range of $p$

digits in $r_n$: $2$

$r_n = a_n \div p$

$10 \le a_n \div p \lt 100$

$\max{(a_n)} \div 100 \lt p \le \min{(a_n)} \div 10$

$\min{(a_n)}=a_4=15$ and $\max{(a_n)}=a_7=95$

$95 \div 100 \lt p \le 15 \div 10$

$0.95 \lt p \le 1.5$

2) Examine the coded results and relate digits in results showing a pattern to each other

The second digit of $c_3, c_5,$ and $c_7$ is 'J'.

$(a_3\div p \mod 10)=(a_5\div p \mod 10)=(a_7\div p \mod 10)$

$(120\div p \mod 10)=(45\div p \mod 10)=(95 \div p \mod 10)$

3) Find a p (within range) that satisfies the relation and produces integers.

$p=1.25$

Paul used a different method involving the highest common factor of the numbers:

All the letters on the right have been increased by the same amount, ie they've been multiplied by the same amount, therefore this number divides all the numbers in the second column. 5 obviously divides all of them but that doesn't give you all 2 digit numbers. If they're divided by 2.5 or $\frac52$ once again it won't result in all 2 digit numbers (15$\div$2.5=6). However if you divide them by 1.25 or $\frac54$ you will achieve all 2 digit integers in column 1. Therefore Column 2 is 125% of column 1 and therefore there is a 25% increase.

**The third table**

Paul used the same highest common factor method:

The answer to part 3 is 40%. The highest common factor of column 2 is 7 but we don't get all 3 digit numbers [if we divide them all by 7]. The only way you can return all the numbers to 3 digit integers is if you divide them by 1.4 or $\frac75.$ Therefore column 2 is 140% of column 1 and therefore the digits have been increased by 40%.

Jeremy used the method of finding upper and lower limits, and considering results with the same last digits:

1) Determine the possible range of $p$

digits in $r_n$: $3$

$\max{(a_n)} \div 1000 \lt p \le \min{(a_n)} \div 100$

$\min{(a_n)}=a_6=147$ and $\max{(a_n)}=a_3=1330$

$1330 \div 1000 \lt p \le 147\div 100$

$1.33 \lt p \le 1.47$

2) Examine the coded results and relate digits in results showing a pattern to each other

The last 2 digits of $c_2$ are 'J'

$(a_2\div p \mod 10)=(a_2\div p \mod 100 - a_2\div p \mod 10)\div10$

$(637\div p \mod 10)=(637\div p \mod 100 - 637\div p \mod 10)\div10$

3) Find a p (within range) that satisfies the relation and produces integers.

$p=1.4$

Ben used a combination of the above methods:

On the 3 digit problem, set the minimum increase again by selecting the largest four digit answer: all the digits in the starting number are different, so the largest number CJH can be is 987, so the minimum percentage increase is 1330$\div$987 = 34.75%. Looking at DDJ, this has a different starting digit to CJH, meaning it has to be less than 900 if CJH is 900 or more: 1239$\div$900 giving a
better lower bound percentage increase of 37.67%. Also, CJH must be greater than 900 if DDJ is between 800 and 900 (it doesn't have to be, but lets assume it is for now, and look beyond this later if we can't find a solution). 1330$\div$900 = 47.78%, so let's look for percentage changes in this window for a solution. 147$\div$1.3767 = 106.8 and 147$\div$1.4778 = 99.5. As 147 divided by the
percentage must be an integer, lets generate some potential solutions by dividing 147 by the integers between these two values: 100, 101, 102, 103, 104, 105 and 106 (using a spreadsheet is quicker). These potential solutions are:

1.47

1.455445545

1.441176471

1.427184466

1.413461538

1.4

1.386792453

Bearing in mind that all the solutions must be integers (and that the problems have had nice round solutions before now!), the best candidate seems to be 1.4. Lets try this and then look at the others if it doesn't work.

In fact, it does work, as dividing each of the end values by 1.4 gives an integer, so the percentage increase was 40%. The full code is:

H=0, E=1, A=3, B=4, J=5, G=6, I=7, D=8, C=9

### Why do this problem?

Code-breaking is often about partial conclusions gradually adding up to possibilities. This problem is unlikely to be done instantly by most students, so discussion should bring up lots of helpful thoughts to share around a group, energising explanation and stimulating individuals into new reasoning and strategy.

### Possible approach

Ask students to look at the code and the column of original values and to share their first thoughts. Hopefully including the insight that whole numbers have stayed as whole numbers after the increase.

The three codings make progressively more demanding challenges.

Maintain an emphasis on the deductive process that establishes the solution rather than merely confirming that a particular multiplier works, though verification should of course be part of the process.

*A teacher comments:**After some initial thought and discussion all (Year 9 set 1) made good progress and found a number of different ways into the problem. The second part of the problem raised points which led neatly into reverse percentages.*

### Key questions

What could all the original numbers be divided by without producing a decimal anywhere in the results column ?

### Possible extension

### Possible support

Encourage exploration to discover the multipliers that tend not to produce many decimal options, and then pick the original numbers so that not even these decimal residuals appear.