Y1T2- IM Math ACE
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This document describes a number trick involving consecutive positive integers between 1 and 100. The trick involves multiplying and manipulating the integers through a series of steps, which always results in the second consecutive integer chosen. The reasoning behind this is explained - no matter which integers are chosen, the operations performed on their product and sums will reduce down to the square of the second integer through cancellation. Zero cannot be included as it would invalidate the trick.
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Y1T2- IM Math ACE
- 1. Mathematics ACE Number Magic! Class: 1A4 Members: Chew Zhen Yuan 1A404 Keefe Ng WS 1A410 S
- 2. Number Trick? S A pattern where: can be applied for a range of results results would be expected
- 3. Number Trick time! S List down 3 consecutive positive integers between 1 and 100 inclusive! (eg 45, 46 and 47) S Take the product of the 3 numbers! (45 x 46 x 47 =97290) S Add 5 to the product! (97290 + 5 = 97295)
- 4. Number Trick time! (continued) S Multiply the result by 10! (97295 x 10 = 972950) S Add another 7 to the product! (972950 + 7 = 972957) S Multiply the sum again by 8! (972957 x 8 = 7783656)
- 5. Number Trick time! (continued) S Take away 456 from the preceding result! (7783656 – 456 = 7783200) S Divide the result by 40! (7783200 ÷ 40 = 194580) S Divide it again by the 2nd consecutive number that you have chosen! (194580 ÷ 46 = 4230)
- 6. Number Trick time! (continued) S Divide the product yet again by 2! (4230 ÷ 2 = 2115) S Add 1 to the result! (2115 + 1 = 2116) S Square root the sum for the final result! √2116 = …… 46!!!!
- 7. Number Trick time! (continued) S No matter which number you try, the result will always be the 2nd consecutive number you have chosen! S Go on, try it with the integers!
- 8. Reasoning S Let’s take the middle integer (consecutive number) as n, shall we? S So, the numbers are: n-1, n, n+1 S If you multiply the products, it will be: (n-1)(n)(n+1)=n3-n
- 9. Reasoning (continued) S If you add 5 to the product… n 3-n+5 S Multiplying it by 10.. 10(n3 – n + 5)= 10n3_10n+50 S After adding 7 to the product… 10n3_10n+50+7 = 10n3_10n+57
- 10. Reasoning (continued) S Multiply the sum by 8: 8(10n3_10n+57) = 80n3-80n+456 S Taking away 456 from the product… 80n3-80n+456-456 =80n3-80n S Dividing it by 40.. (80n3-80n)÷ 40 = 2n3-2n
- 11. Reasoning (continued) S Now, divide it by the second consecutive number.. (2n3-2n) ÷ n = 2n2-2 S Divide it by 2! (2n2-2) ÷ 2 = n2-1 S Add 1 to the result… n2-1+1 = n2
- 12. Reasoning (continued) S Lastly, square root the sum! √n2= n (the 2nd consecutive number!!) S Do you see the magic now? If you don’t, go back and analyze the steps. Maybe you went wrong somewhere?
- 13. Question! Qns: Can zero be included? Ans: No Exp: 0 is a neutral integer. If it is included, the product of the three numbers would be regarded as the product of two. As a result, if the integers are taken as 0, 1 and 2, the final answer will be the square root of 3, which is 1.732050808…. Wrong!!
- 14. Oh wait!!!! S Didn’t like the presentation?? S Here’s a potato