course Mth 151 y?????y????assignment #008
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00:02:00 `q001. There are seven questions in this set. See if you can figure out a strategy for quickly adding the numbers 1 + 2 + 3 + ... + 100, and give your result if you are successful. Don't spend more than a few minutes on your attempt.
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RESPONSE --> You can pair 1 with 100, 2 with 99, 3 with 98.... There are 50 pairs of numbers, since half of 100 is 50. Those pairs always are 101. So you take 50 * `101 to get 5050 confidence assessment: 3
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00:02:06 These numbers can be paired as follows: 1 with 100, 2 with 99, 3 with 98, etc.. There are 100 number so there are clearly 50 pairs. Each pair adds up to the same thing, 101. So there are 50 pairs each adding up to 101. The resulting sum is therefore total = 50 * 101 = 5050.
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RESPONSE --> OK self critique assessment: 3
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00:03:00 `q002. See if you can use a similar strategy to add up the numbers 1 + 2 + ... + 2000.
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RESPONSE --> There are first 1000 pairs because half of 2000 is 1000. You can add 2001 + 1..... So you take the 1000 pairs and multiply them by 2001 to get 2,001,000. confidence assessment: 3
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00:03:06 Pairing 1 with 2000, 2 with 1999, 3 with 1998, etc., and noting that there are 2000 numbers we see that there are 1000 pairs each adding up to 2001. So the sum is 1000 * 2001 = 2,001,000.
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RESPONSE --> OK self critique assessment: 3
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00:06:00 `q003. See if you can devise a strategy to add up the numbers 1 + 2 + ... + 501.
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RESPONSE --> Like the previous problems, you can add 501 to 1..... Every pair gives you 502 except one of the numbers in the middle. Take the one number out of 501 and it gives you 500. The middle number is 266. You then take 502 * 250 + 266 to get 125,751. confidence assessment: 2
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00:06:08 We can pair 1 with 501, 2 with 500, 3 with 499, etc., and each pair will have up to 502. However there are 501 numbers, so not all of the numbers can be paired. The number in the 'middle' will be left out. However it is easy enough to figure out what that number is, since it has to be halfway between 1 and 501. The number must be the average of 1 and 501, or (1 + 501) / 2 = 502 / 2 = 266. Since the other 500 numbers are all paired, we have 250 pairs each adding up to 502, plus 266 left over in the middle. The sum is 250 * 502 + 266 = 125,500 + 266 = 125,751. Note that the 266 is half of 502, so it's half of a pair, and that we could therefore say that we effectively have 250 pairs and 1/2 pair, or 250.5 pairs. 250.5 is half of 501, so we can still calculate the number of pairs by dividing the total number of number, 501, by 2. The total sum is then found by multiplying this number of pairs by the sum 502 of each pair: 250.5 * 502 = 125,766.
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RESPONSE --> OK confidence assessment: 3
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00:10:43 `q004. Use this strategy to add the numbers 1 + 2 + ... + 1533.
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RESPONSE --> I did what I did earlier, and added the pairs. Each pair equals to 1534. Half of 1534 is 767. That's the left over number in the middle. So you take 1534 times 766.5 + 767 to get 1176578. confidence assessment: 2
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00:11:09 Pairing the numbers, 1 with 1533, 2 with 1532, etc., we get pairs which each adult to 1534. There are 1533 numbers so there are 1533 / 2 = 766.5 pairs. We thus have a total of 1534 * 766.5, whatever that multiplies out to (you've got a calculator, and I've only got my unreliable head).
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RESPONSE --> I was using similar tactics to the previous problem, and I'm confused why one can use 766.5 and not 767. self critique assessment: 2
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00:14:07 `q005. Use a similar strategy to add the numbers 55 + 56 + ... + 945.
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RESPONSE --> I paired up 55 and 945.... Those gave me 1000 for every pair. I subtracted 55 from 945 yet added 1 to find the unit jumps. That gave me 891 for the sum. If we have 891 numbers in the sum, we have 891/2 = 445.5 pairs, each adding up to 1000. That gives us 445,500. confidence assessment: 2
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00:14:12 We can pair up 55 and 945, 56 and 944, etc., obtaining 1000 for each pair. There are 945 - 55 + 1 = 891 numbers in the sum (we have to add 1 because 945 - 55 = 890 tells us how many 1-unit 'jumps' there are between 55 and 945--from 55 to 56, from 56 to 57, etc.. The first 'jump' ends up at 56 and the last 'jump' ends up at 945, so every number except 55 is the end of one of the 890 'jumps'. But 55 is included in the numbers to be summed, so we have 890 + 1 = 891 numbers in the sum). If we have 891 numbers in the sum, we have 891/2 = 445.5 pairs, each adding up to 1000. So we have a total of 445.5 * 1000 = 445,500.
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RESPONSE --> OK self critique assessment: 3
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00:16:40 `q006. Devise a strategy to add the numbers 4 + 8 + 12 + 16 + ... + 900.
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RESPONSE --> Pairing 4 with 900... The difference between 4 and 900 is 896. 896/4 gives us 224 jumps, so I added one to give me 225 cause it doesn't end at the first. 225 / 2 = 112.5 pairs that add up to 904, and that total is 101,700. confidence assessment: 2
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00:16:45 Pairing 4 with 900, 8 with 896, etc., we get pairs adding up to 904. The difference between 4 and 900 is 896. The numbers 'jump' by 4, so there are 896 / 4 = 224 'jumps'. None of these 'jumps' ends at the first number so there are 224 + 1 = 225 numbers. Thus we have 225 / 2 = 112.5 pairs each adding up to 904, and our total is 112.5 * 904.
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RESPONSE --> OK self critique assessment: 2
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00:17:38 `q007. What expression would stand for the sum 1 + 2 + 3 + ... + n, where n is some whole number?
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RESPONSE --> I believe it is n/2 *n+1 confidence assessment: 2
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00:17:43 We can pair 1 and n, 2 and n-1, 3 and n-2, etc., in each case obtaining a sum of n + 1. There are n numbers so there are n/2 pairs, each totaling n + 1. Thus the total is n/2 * (n+1).
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RESPONSE --> OK self critique assessment: 3
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