#$&* course Mth 151 6/16 3:46 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: These numbers can be paired as follows: 1 with 100, 2 with 99, 3 with 98, etc.. There are 100 number so there are clearly 50 pairs. Each pair adds up to the same thing, 101. So there are 50 pairs each adding up to 101. The resulting sum is therefore total = 50 * 101 = 5050. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): After viewing the given solution, I now see an easy way to add all of these numbers up. I didn’t even realize that pairing the numbers in such a manner would give you the solution but now I know. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q002. See if you can use a similar strategy to add up the numbers 1 + 2 + ... + 2000. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Begin by pairing the numbers ex. 1,2000 or 2 and 1999 one would always get 2001 so 1000 * 2001 would be 2,001,000 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q003. See if you can devise a strategy to add up the numbers 1 + 2 + ... + 501. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: One would you the same strategy ex. 1, 501 or 2 and 500 etc. would give us 502 so 250.5 * 502 would be 125,751. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We can pair 1 with 501, 2 with 500, 3 with 499, etc., and each pair will have up to 502. However there are 501 numbers, so not all of the numbers can be paired. The number in the 'middle' will be left out. However it is easy enough to figure out what that number is, since it has to be halfway between 1 and 501. The number must be the average of 1 and 501, or (1 + 501) / 2 = 502 / 2 = 251. Since the other 500 numbers are all paired, we have 250 pairs each adding up to 502, plus 251 left over in the middle. The sum is 250 * 502 + 251 = 125,500 + 251 = 125,751. Note that the 251 is half of 502, so it's half of a pair, and that we could therefore say that we effectively have 250 pairs and 1/2 pair, or 250.5 pairs. 250.5 is half of 501, so we can still calculate the number of pairs by dividing the total number of number, 501, by 2. The total sum is then found by multiplying this number of pairs by the sum 502 of each pair: 250.5 * 502 = 125,766. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q004. Use this strategy to add the numbers 1 + 2 + ... + 1533. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the same strategy 1 + 1533 or 2+1532 etc. would make 1534 so 766.5 * 1534 would total 1,175,811. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q005. Use a similar strategy to add the numbers 55 + 56 + ... + 945. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the same strategy 55+945 or 56+944 etc. = 1000 so 1000*445.5< (945-54/2) The total would be 445,500. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We can pair up 55 and 945, 56 and 944, etc., obtaining 1000 for each pair. There are 945 - 55 + 1 = 891 numbers in the sum (we have to add 1 because 945 - 55 = 890 tells us how many 1-unit 'jumps' there are between 55 and 945--from 55 to 56, from 56 to 57, etc.. The first 'jump' ends up at 56 and the last 'jump' ends up at 945, so every number except 55 is the end of one of the 890 'jumps'. But 55 is included in the numbers to be summed, so we have 890 + 1 = 891 numbers in the sum). If we have 891 numbers in the sum, we have 891/2 = 445.5 pairs, each adding up to 1000. So we have a total of 445.5 * 1000 = 445,500. STUDENT COMMENT I got very confused on this one. I don’t quite understand why you add a 1. INSTRUCTOR RESPONSE For example, how many numbers are there in the sum 5 + 6 + 7 + ... + 13 + 14 + 15? 15 - 5 = 10. However there are 11 numbers in the sum (5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q006. Devise a strategy to add the numbers 4 + 8 + 12 + 16 + ... + 900. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using a similar strategy 4 +900 or 8 + 896=904 900/8= 112.5 so the sum would be 101,700 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q007. What expression would stand for the sum 1 + 2 + 3 + ... + n, where n is some whole number? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: n/2 * (n+1) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!