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 Grade 10 to 12 Funny Factorisation Firstly think about all possible solutions for the top row. It cannot be a number above 250 or below 100 and it has to end in a 9. The top number cannot have a 4 or a 6 or another 9. The only possibilities are 129, 139, 159, 179, 189, 219 and 239. So try these numbers with every 2 digit number beginning with a 4 until you find the answer 159 x 48 = 7632. The next funny factorization is: 186 x ** = **** and the remaining digits are 2, 3, 4, 5, 7 and 9. The largest possible value of **** is 9754 and 9754/186 = 52.4 (to one place of decimals). So the 2-­‐digit number we are looking for is between 23 and 52. The possibilities are 23, 24, 25, 27, 29, 32, 34, 35, 37, 39, 42, 43, 45, 47, and 49. We have 15 cases to check. The solution is 186 x 39 = 7254 To split 4396 into a product of a 3-­‐digit and a 2-­‐digit number first find the prime factorisation of 4396. 4396 = 2 x 2198 = 2 x 2 x 1099 = 2 x 2 x 7 x 157 The only ways to make this into a product of a 3-­‐digit and 2-­‐digit number are: 314 x 14 or 157 x 28. The solution (which must uses all the digit 1 to 9) is 4396 = 157 x 28. The number 5796 can be written as the product of a 3-­‐digit number and a 2-­‐digit number in the form *** x ** using the digits 1 to 9 each once and only once in two DIFFERENT ways. First factorise 5796 into prime factors: 5796 = 2 x 2898 = 2 x 2 x 1449 = 2 x 2 x 3 x 483 = 2 x 2 x 3 x 3 x 161 2
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= 2 x 2 x 3 x 3 x 7 x 23 = 2 x 3 x 7 x 23 The remaining digits to use are 1, 2, 3, 4 and 8. To get 6 as the units digit in the product the units digits must be 2 & 3 or 2 & 8. The 2-­‐digit factors ending in 2, 3 or 8 can be: 12, 13, 18, 23, 28, 32, 38, 42, 43, 48, 82 and 83 Dividing 5796 by these numbers to test these cases gives the solutions: 483 x 12 = 5796 and 138 x 42 = 5796. 5
Similarly 5346 = 2 x 3 x 11 and the funny factorisations are: 5346 = 198 x 27 and 5346 = 297 x 18. We have found seven solutions. Notes for Teaching Why do this problem This problem reinforces the understanding of place value, it gives learners practice in finding prime factors, and it enables them to do something with the prime factors. We teach learners to decompose numbers into prime factors and seldom give them any chance to put this to use. It is important to know how prime numbers form the building blocks of all whole numbers and to know that all whole numbers can be split into a product of prime factors (The Prime Factorization Theorem). This theorem is fundamental to arithmetic. Learners need a good understanding of the processes of arithmetic to enable them to do algebra. Understanding how to factorise whole numbers helps learners later on to factorise algebraic expressions. Funny Factorisation is seven problems in one with each problem offering a new and slightly different challenge to those who work faster. Solving the first two problems calls for an understanding of place value and estimating the sizes of the factors. The last five problems require factorization of numbers into their prime factors. With each problem, learners need to work systematically to list possible cases and to exclude others. The number of possible cases, that is the number of calculations to try out, can be narrowed down using some clear thinking and estimation of the magnitudes of certain products. Suggestions for teaching: Before starting on this problem check that learners know how to find the prime factors of a given number. You might give some help with the first problem so that the learners gain confidence but the teacher should try to encourage learners to refuse help and to want to solve the problem on their own. Let the learners use calculators. Working in pairs the learners could take turns, one using the calculator for a few minutes while the other writes down the results. In any problem solving activity the first step is to understand the problem so make sure that the learners know that the digits they must use to replace the * symbols in the first problem * * 9 x 4 * = * 6 * * are 1, 2, 3, 5, 7 and 8 (not 4, 6 and 9 which are used already). As a hint, you might ask them “what is the smallest 2-­‐digit number and what is the biggest 3-­‐digit number we can have in this multiplication?” A good answer here is “41 is the smallest 2-­‐digit number and 250 is the biggest 3 digit number”. Another possible answer is that the biggest 4 digit number is 8675 and by dividing this by 41 we can tell that the 3-­‐digit number is less than 211. Either way we know that the 3-­‐digit number is between 100 and either 211 or 250, and we know it ends in a 9, so there are only a few possibilities. Ask the learners to make a list of all the possible 3 digit numbers and then try to find the solution. You might use the 1 – 2 – 4 – more strategy here, giving some time for learners to work on this problem individually, then in pairs, then in fours. When they have solved the first problem suggest that the groups try to think about the second one in a similar way. The third problem comes with a hint to find the prime factors of 4396 and all the remaining problems use prime factorization. Learners will almost certainly try different possibilities in some random order. Let them work on the problem for a while and then ask them if they can save themselves some work by thinking up a good system to rule out the factors which can’t be used and for checking that they have found all the solutions. Learners should be praised for giving good explanations and it is not necessary for everyone to try all seven examples. Key Questions 1. Which digits can still be used? 2. What are the biggest (or smallest) 4-­‐digit numbers we can have in this multiplication? 3. What is the biggest (or smallest) 2-­‐digit number we can have in this multiplication? 4. What is the biggest (or smallest) 3-­‐digit number? 5. Can you make a list of all the possibilities? 6. Are there some digits we know we can’t use as units digits? 7. What are the prime factors of that number? 8. Can you make 2 groups of prime factors to give a 2-­‐digit and a 3 digit number which don’t repeat any digits? Possible Support If a learner is really struggling you could supply one or two of the missing digits and ask the learner to try to find the rest. The problem One to Eight http://nrich.maths.org/757 is similar and a bit easier than Funny Factorisation. Possible Extension The teachers role is to try to help learners to be more independent and not to ‘spoon feed’ them. Only offer hints to the learners who really need support. The key questions listed here give hints but as your learners do more problem solving and develop skills and confidence they will need less help from the teacher. Try FOOT+BALL=SUM http://nrich.maths.org/576