A Change in Code

4799 /www/nrich/html/content/id/4799/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> All numbers in the first column have been increased by the same percentage to give the results shown in the second column.<div></div> <div></div> But the result numbers are disguised - their digits are replaced by letters. A given letter stands for the same numeral every time it appears in that column. For example, whatever the C stands for, it stands for that numeral every time the C appears. Can you work out the percentage change, even with the result digits in code? <div></div> <div class="editorial"></div> <div class="editorial"><mdo:image width="268" height="376" alt="a change in code (first)" src="A%20Change%20in%20Code%20%28FIRST%29.JPG"></mdo:image></div> <div></div> What about this next set? It's a new coding - the letters now stand for different numerals. What is the percentage change used on all numbers in the first column this time? <mdo:image width="262" height="379" src="A%20Change%20in%20Code%20%28SECOND%29.JPG" alt=""></mdo:image> And here? Different coding again, and much harder. By now the problem has definitely become a three star challenge - fancy having a crack at it? <mdo:image width="283" height="376" src="A%20Change%20in%20Code%20%28THIRD%29.JPG" alt=""></mdo:image> </mdoxml> <mdoxml version="1.0">Good thinking from several people on this problem. Below is a combination of solutions from Paul at Castleknock College, and Jean in Warwick Â 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 be 99 or else there would be 2 equal letters. All numbers have one thing in common - they're divisible by 5 So the only way all of these numbers could produce integers, with no fraction or decimal part, is if their increase is calculated using fifths. It must be 20% because anything greater, 40% 60% etc., would turn 80 into a three digit number which it clearly isn't. An alternative explanation could be : 80 to HF is an increase but still only produces a 2 digit number, so HF is 81 to 99. Trying each percentage, for example 80 to 90 is 12.5%, all give Â results Â involving fractions from some of the start values, except for 80 to 96 (20%). And 20% checks as consistent across all data. In Part 2 the answer is 25% BC to 15 so B is 1 and C is 2, 3, or 4 CG to 35 so C is in fact 2 or 3 Â C is not zero so BC is either 11, 12, or 13 11 to 15 and 13 to 15 do not give a percentage that keeps all other results integer. 12 to 15 is an increase of 25% and this is consistent for all the other data. In Part 3. the percentage increase was 40% Increasing a number is like multiplying with a top-heavy fraction. For example a 5% increase is achieved multiplying the original number by 21/20. Dividing by 21/20 will reverse that process, and multiplying by 20/21 is the same thing. Â I notice that all the results contain a factor of 7, so I am looking for a/7 as that multiplier, where a< 7, and choosing a so as to avoid fractions as required. EHJ goes to 147 So EHJ is between 101 and 146, because it has 3 digits and this is an increase. If EHJ/147 is equivalent to the fraction a/7 then EHJ is 105 or 126. Giving an a value of 5 or 6 Next I bring in CJH which increases to 1330 1330 multiplied by 6/7 is 1140, which is four digits not three, so 5/7 is the correct multiplier. The percentage increase was 40% </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h3>Why do this problem?</h3> Code-breaking is often about partial conclusions gradually adding up to possibilities. <a href="http://nrich.maths.org/public/viewer.php?obj_id=4799">This problem</a> 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. <h3>Possible approach</h3> 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. <h3>Key questions</h3> <ul> <li>What could all the original numbers be divided by without producing a decimal anywhere in the results column ?</li> </ul> <h3>Possible extension</h3> <div>Able students may like to design similar coded columns for each other to crack.</div> <div>Discussion may include an exploration of how many values need to be seen coded before the solution multiplier is known for sure.</div> <h3>Possible support</h3> <div>Students who are not ready for this problem without some preliminary activity should generate simple two-digit 'codings' for themselves and swap these around the group for others to crack.</div> <div>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.</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">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? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Part 1. 80 to HF is an increase but still only produces a 2 digit number, so HF is 81 to 99. Trying each percentage, for example 80 to 90 is 12.5%, all give fractional results from some of the start values, except for 80 to 96 (20%). And 20% checks as consistent across all data. Part 2. BC to 15 so B is 1 and C is 2, 3, or 4 CG to 35 so C is in fact 2 or 3 C is not zero so BC is either 11, 12, or 13 11 to 15 and 13 to 15 do not give a percentage that keeps all other results integer. 12 to 15 is an increase of 25% and this holds consistent with all other data. Part 3. All the results contain a factor of 7 So reversing the percentage change by multiplying by a/7 where a< 7 would avoid fractions as required. EHJ goes to 147 So EHJ is between 101 and 146, and if EHJ/147 is to reduce to a/7 then EHJ is 105 or 126. Giving an a value of 5 or 6 Now using CJH to 1330 1330 multiplied by 6/7 is 1140 and CJH is three digits only, so 5/7 is the correct multiplier. The percentage increase was 40% </mdoxml> 2 3 0 0 0 1 0 A Change in Code There are two sets of numbers. The second is the result of the first after an increase by a constant percentage. How can you find that percentage if one set of numbers is in code? Fractions, Decimals, Percentages, Ratio and Proportion Mixed fractions, decimals and percentages Admin Long problems Fractions, Decimals, Percentages, Ratio and Proportion Calculating with percentages