Wellington Mathematics Association

1. Wellington Mathematics Association April 21 2010

3. Divisibility • Problem-Solving with HCF and LCM • Problem-Solving with Graphs • Solving Diophantine Equations with Modulo Arithmetic

4. Divisibility by 11 What is the remainder when 6987 is divided by 11? Remainder = (7+9) – (6+8) = 2 i.e. Sum of Odds – Sum of Evens, starting from the left.

5. Further Example What is the remainder when 725391 is divided by 11? Remainder = Odds - Evens = (1+3+2) – ( 9+5+7) = -15 = -4 = 7 i.e. Add multiples of 11 until you get satisfactory answer.

6. Divisibility by 99 What is the remainder when 2684 is divisible by 99? Remainder is 84 10 +26+01 110 11 Remainder = 11.

7. Further Example What are the values of a and b if 14ab8 is divisible by 99? The addition b8 4a + 01 has to give 99 or 198 etc which means a = 0 and b = 5, making the number 14058 divisible by 99.

8. If Janet can paint a house in 6 days while • John takes 9 days to paint the same • house, how long will it take them if they • paint it together?

9. Now 18 is the LCM of 6 and 9 and is the smallest number of days in which they would each paint the house a whole number of times. In fact Janet paints it 3 times, while John paints it twice, so working together they would paint it 5 times in 18 days, and therefore they would complete it once in 3.6 days

10. The notion can be extended to 3 participants, or any number for that matter. The question can also be posed in reverse. If two people can set a table in 5 minutes and Jane can do it by herself in 8 minutes, then how long would it take Jacob?

11. Consider what would happen in 40 minutes. They could set the table 8 times between them, but Jane could do it 5 times by herself, leaving Jacob setting the table 3 times in 40 minutes i.e. once in 40/3 minutes.

12. Other contexts where this sort of thinking can apply are: • Bath filling with water from both hot and cold taps, with and without the plug in. • Two trains travelling simultaneously on the same line • Two swimmers simultaneously doing laps in a pool at different speeds

13. Now let us turn our attention to problem-solving with graphs. Note that I do not use a single equation, nor do I mention the words gradient and intercept. In other words this is basically drawing pictures of situations and employing a little intuition in places.

14. Now let us look at a problem involving two candles. Two candles of the same length are lit simultaneously. If one candle takes 8 hours to burn out, and the other takes 10 hrs to burn out, when will the slower-burning candle be twice the length of the faster-burning one? ?

17. The graphs show us the length of the slower-burning one is twice the length of the faster-burning one after about 6 and 2/3hrs. Other contexts include 2 runners on the 400m track, and when one is twice as far from finish as the other.

18. Now let us have a look at how those LCM problems can be solved graphically. If Johnny can paint a house in 10 days while Peter could paint the same house in 15 days, how long would it take them if they worked together?

19. We use ideas very similar to the previous one, but one of the graphs shows howmuch one of the men has done, while the other shows how much theother has left. When one has completed what the other still has to complete, then they have painted the house between them.

21. After 6 days, the work done by Johnny equals the work still to be done by Peter. Therefore they have done the paint job between them.

23. Let us now look at solving some mixture problems using graphs. In this work • all the combinations that satisfy a given ratio form a line • The actual amount represented by a point is the sum of the two components at that point

24. Let us look at one ratio of, say, vinegar and water, in the ratio 2:1.

26. The graph can be used to add the coordinates of a point by passing an ‘x + y’ line through the point. Through the point (4,2), this line cuts both axes at 6, thus giving us the total of 4 and 2.

28. Using these ideas let us solve the following mixture problem: If I have a 2:1 mixture of vinegar and water and a 4:5 mixture of vinegar and water, how should I combine these mixtures to obtain 18 mLs of a 1:1 mixture?

31. We have added the demand line for 18 mLs of 1:1 mixture, thus isolating the critical point (9,9)

32. Now we perform a ‘vector’ addition of the two available ‘vectors’, so that their sum is the ‘vector’ (9,9). We draw a line through (9,9) parallel to the 1:1 line until it reaches the 4:5 line

34. This gives vector OH, representing the quantity of 4:5 mixture which is needed, and vector HG which represents the quantity of 2:1 mixture which should be used.

35. We now use the ‘x + y’ line through H to work out the actual number of mLs of 4:5 mixture to be used. The sum will be the intercept value on either axis, while the number of mLs of 2:1 mixture will be what is required to make up the 18 mLs.

37. The graph shows that to achieve 18mLs of 1:1 we need to mix 13.5 mLs of the 4:5 mixture with 4.5mLs of the 2:1 mixture.

38. To solve a diophantine equation is to solve a single equation in two or more variables for integer solutions. One method of solving them is to use modulo arithmetic, which essentially is remainder arithmetic. I limit the work to linear equations.

39. Solve 11x +24y = 271 for positive integer solutions. Looking at remainders of both sides after division by 11, we get 0 + 2y = 7 2y = 7,18, 29, 40,… y = 9, 20, ….. Backsubbing y = 9, gives x = 5, so that (5,9) becomes the only solution in the positive integers

40. Here is another example. A retailer sells two types of drills, type A for $99, and type B for $119. If weekly receipts for sales of both types is $12450, how many of each type are sold?

41. 99a + 119b = 12450 After div by 99 0 + 20b = 75 20b = 75 + 5 x 99 = 570xxx 20b = 75 + 15 x 99 = 1560 b = 78, giving a = 32 The retailer sold 32 of type A, and 78 of type B.

42. A Lucas sequence is a sequence whose terms are formed by the addition of the two previous ones. The celebrated Fibonnaci sequence starts with 1,1, 2, 3,… The question is: Find the Lucas sequence for which the 11th term is 2010?

43. If the first two terms are a, b, then the sequence is a, b, a+b, a+2b, 2a+3b, etc with 11th term equal to 34a + 55b. 34a + 55b = 2010 In mod 34 0 + 21b = 4, 38, 72, 106, 140, 174, 208. Since -13 = 21 in mod 34, 208/-13 = -16, which equals 18. b = 18, a = 30, giving the sequence 30, 18, 48, 66, 114, 180, 294, 474, 768, 1242, 2010,