course Mth 151 ӊeϾassignment #008
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21:37:56 `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 could use the method of Gauss. 1+100= 101 2+99=101 3+98=101 So you have 50 sums of 101 50*101=5050 confidence assessment: 3
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21:38:04 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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21:39:42 `q002. See if you can use a similar strategy to add up the numbers 1 + 2 + ... + 2000.
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RESPONSE --> 1+2000= 2001 2+1999= 2001 3+1998= 2001 So there are 1000 sums of 2001 1000*2001= 2,001,000 confidence assessment: 3
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21:39:47 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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21:41:10 `q003. See if you can devise a strategy to add up the numbers 1 + 2 + ... + 501.
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RESPONSE --> 1+501= 502 2+500= 502 3+499= 502 So there are 250.5 sums of 502 250.5*502= 125,751 confidence assessment: 2
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21:43:36 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 --> 250.5*502 is 125, 751 not 125, 766 confidence assessment: 2
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21:45:38 `q004. Use this strategy to add the numbers 1 + 2 + ... + 1533.
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RESPONSE --> 1+ 1533= 1534 2+ 1532= 1534 3+ 1531= 1534 So we have 766.5 pairs of 1534 766.5*1534= 1,175,811 confidence assessment: 3
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21:45:51 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 --> OK self critique assessment: 3
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21:47:07 `q005. Use a similar strategy to add the numbers 55 + 56 + ... + 945.
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RESPONSE --> 55+945= 1000 56+944= 1000 57+943= 1000 We have 472.5 pairs of 1000 472.5*1000= 472,500 confidence assessment: 3
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21:47:59 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 --> I forgot to subtract 55 from 945 to obtain the total numbers in the sum self critique assessment: 3
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21:51:39 `q006. Devise a strategy to add the numbers 4 + 8 + 12 + 16 + ... + 900.
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RESPONSE --> 4+900= 904 8+896= 904 12+892= 904 16+888= 904 Divide 900 by 4 to find the number of pairs. 900/4= 225 There are 225 pairs of 904 225*904= 203, 400 confidence assessment: 2
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21:52:43 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 --> After I divided the number by 4, I forgot that I needed to divide the number again in order to find the number of pairs self critique assessment: 2
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21:54:04 `q007. What expression would stand for the sum 1 + 2 + 3 + ... + n, where n is some whole number?
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RESPONSE --> 1+2+3+.....+n= n(n+1)/2 confidence assessment: 2
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21:55:26 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 --> I believe that this obtains the same results as the formula I used self critique assessment: 2
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