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Dr J Frost (jfrost@tiffin.kingston.sch.uk)
KS3 Number Grids Dr J Frost Last modified: 8th December 2014
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Starter In the Maths Challenges and Olympiads, you are often asked to insert the numbers 1 to π into a structure such that the sum of each row/column/line in the structure is the same. Q The numbers 1 to 9 all appear once in the following structure. The sum of each line is the same. What are the missing numbers? 1 3 5 6 4 9 ? ? ? ? 8 7 2 ? ?
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General Tips Q The numbers 1 to 9 all appear once in the following structure. The sum of each line is the same. What are the missing numbers? Bro Tip 1: Calculate the sum of 1 to π. The quick way is 1 2 π π+1 1 3 1 2 Γ9Γ10=9Γ5=45 ? 7 Bro Tip 2: Let π say be the total in each line. Form an equation, one side in terms of π and the other side in terms of numbers used (considering overlap) ? 4π= (if we add the 4 lines, 1, 7 and 3 are each seen twice) 4π=56 π=14 This then makes the problem much easier, as the middle two numbers are now obvious, and thereβs little to experiment with in the outer boxes.
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More Tips Q2 Again, the numbers 1 to 9 appear in each box. Let π» be the total of each row. Find the possible values of π» Bro Tip 3: Give any overlapping squares a (separate) variable name. 1 3 ? ο See diagram. Form an equation, one side in terms of π and the other side in terms of numbers used (considering overlap) π₯ ? 4π= π₯ 4π=49+π₯ Bro Tip 4: Reason about divisibility on each side of the equation. ? LHS is divisible by 4, so RHS must be. Thus π₯=3, 7. But 3 already used, so π₯=7, and thus π=14
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Final Tip Q3 Again, the numbers 1 to 9 appear in each box. Let π» be the total of each row. Find the possible values of π» Give any overlapping squares a (separate) variable name. π¦ π₯ ? ο See diagram. 1 Form an equation, one side in terms of π and the other side in terms of numbers used (considering overlap) Bro Tip 5: Consider the smallest and largest your variables could be to give a plausible range for π. 4π=45+1+π₯+π¦ 4π=46+π₯+π¦ ? What would we do at this point? The smallest π₯ and π¦ could be is 2 and 3 (1 is used) 4π= =51 β π=12.75 So π is at least 13. The largest they could be is 8 and 9. 4π= =63 β π=15.75 So π is at most 15. i.e. 13β€πβ€15 ? ? Weβve only shown π can plausibly be between 13 and 15, but we havenβt shown these values actually work. Consider each value of π in turn and show it can lead to a valid arrangement.
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Olympiad Worksheet Worksheet provided. Here is a reminder of the tips:
Full solutions on next slides. Bro Tip 1: Calculate the sum of 1 to π. The quick way is 1 2 π π+1 Bro Tip 3: Give any overlapping squares a (separate) variable name. Bro Tip 2: Let π say be the total in each line. Form an equation, one side in terms of π and the other side in terms of numbers used (considering overlap) Bro Tip 4: Reason about divisibility on each side of the equation. Bro Tip 5: Consider the smallest and largest your variables could be to give a plausible range for π.
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J25 ?
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J38 ?
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J24 ?
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J48 ?
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M30 ?
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M48 ?
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M66 ?
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M72 ?
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M65 ?
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