POTW #11 Grade 8 Question:
POTW #11 Grade 7 Question:
POTW #10 Grade 8 Answer:
POTW #10 Grade 7 Answer:
Thursday, November 17, 2016
POTW #11 - The Challenge!
This week it was so great to see students challenging themselves with a more difficult question and asking their peers for support when needed. Always remember that we have a few class leaders that can offer support if you need it!
POTW #11 Grade 8 Question:
POTW #11 Grade 7 Question:
POTW #10 Grade 8 Answer:
POTW #10 Grade 7 Answer:
POTW #11 Grade 8 Question:
POTW #11 Grade 7 Question:
POTW #10 Grade 8 Answer:
POTW #10 Grade 7 Answer:
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Grade 7 POTW:
ReplyDeleteHow I did it was to calculate a possibility for each number of numbers.
Starting from 2 numbers, I can see that it is impossible because you can't have 499+500 or 500+501. These do not equal to 1000, and I know the numbers have to be consecutive.
Now, I go to three numbers. I know this is also impossible because 1000 is not divisible by 3, so I cannot average it out into 3 consecutive values.
Coming to four numbers, I see that 1000 is divisible by it, which gives me 250. But, because the median would have to be 250, there is no way that I can get 1000 with 4 numbers.
For example, if I tried 248+249+250+251 that would be 999, and if I tried 249+250+251 that would be 1001. Therefore, 4 numbers would not work.
Now, I go to 5 numbers. Divided, the quotient is 200. 200 would be the median, and the rest also make a mean of 200 since I can have 201 and 202 on one side, and 199 and 198 on the other side, adding up to 1000.
Using all of this calculations, my answer for the POTW is 5 numbers needed.
--Alan--
Alan I think you messed up the addition for when there are 4 consecutive numbers. 248 + 249 + 250 + 251 = 998, not 999 and you only put the numbers for the other option. I think that you were trying to write 249+250+251+251 but that equals 1002, not 1001. But regardless, 4 consecutive numbers do not work.
DeleteOops! Small mistakes. THX AVI!
DeleteHere's how I solved it:
ReplyDeleteI used guess and check along with some math XD:
So first I was stumped like " THIS IS GONNA TAKE SO LONG!!!" And the I thought and calmed down, " Why don't I just break this up into pieces?" So that's what I did. I first found out consecutive numbers that added up together to equal a sum with the ones digit being zero ( due to the total having to be 1000), After that I tested to see what consecutive numbers can add up to near 1000, with the lowest amount used, I found my range of numbers at 190-210. Then I tested my method out to see how close to 1000 I was.
Example:
6 + 7 + 8 + 9 + 0 = 30
196 + 197 + 198 + 199 + 200 =
990
After nine tries this is what I got:
8 + 9 + 0 + 1 + 2 = 20
198 + 199 + 200 + 201 + 202 =
1000
How do I know that this is the lowest?
No consecutive numbers that used under 5 consecutive numbers added up to a value with the one digit 0 ( at least when I tried).
The smallest amount of positive integers is 5
POTW 10:
Uh oh... I totally forgot! Q~Q
Looking at the question...
this is really hard ._.
So after drawing out a visual aid, I noticed that the hexagon was made up of two rectangles, one big isosceles triangle and two smaller ones. So I found the area of the two rectangles ( 100cm^2 when you add both of the areas together) then found the area of the two small isoceles triangles and than the bigger one. I added it together and got 355cm^2, off by around 120cm squared XD. This question was difficult due to the ISOSCELES TRIANGLES EVERYWHERE. ^_^ But At least I tried even if it's late. what I did wrong? Mostly like calculated the area of the big isosceles triangle wrong or somwthing XD
Grade 7 POTW:
ReplyDeleteI know instantly that 2 and 3 won't work because 3 isn't a factor of 1000 and 2 won't meet the criteria because 499+500=999, but when you evaluate 500+501, you know that it is either 1 digit below or higher than 1000. When I got to 4, I know that 4 is a factor BUT it also didn't meet the criteria, like 2, it was either one digit up, or below 1000. Once we get to 5 digits, I know this might work because 5 is a factor of 1000. Since 5*200=1000, using the guess and check method, I come up that 5 numbers will be the minimum amount with 198+199+200+201+202.
For the POTW 10, I'm doing it right now on the other page. Also, Alan you have a little error where you put 3 numbers instead of 4 for your POTW.
I LOVE the POTW #10 Gr. 7 question by the way. Do you see how and why it is 10.9cm? So many steps and what an elaborate problem!
ReplyDeleteI got an answer of 28 as the smallest number, as I used the average for 25 numbers.
ReplyDeleteMy strategy to find numbers that add consecutively to sum a number was to divide 1000 by a factor and create a data set where the quotient is the median and there are the amount of numbers that I divided 1000 by.
ReplyDeleteI started with 2 numbers as that is the minimum and I got a quotient of 500. No consecutive numbers can have the median of 500 so I need to move on.
3 doesn't work because 3 is no a multiple of 1000.
If I do 4 numbers, the median of those numbers must be 250(quotient). No 4 consecutive numbers can have the median of 150.
5 numbers work because those numbers must have a median of 200. Those 5 numbers are 198, 199, 200, 201, 202.
198 + 199 + 200 + 201 + 202 = 1000 <---works.
The minimum number of consecutive integers that sum to 1000 is 5.
Avi, for your fourth paragraph, second sentence, do you mean 250 instead of 150
DeleteIt would take 5 consecutive integers being 198, 199, 200, 201, and 202. I got this because the low multiples of 1000 are 2, 4, 5, 8 and 10. Going down the list it's impossible to do 2 because then it would be either 999 or 1001. Then it can't be 4 because it would be 998 or 1002. Then 5 would work being 198, 199, 200, 201 and 202 with each side balancing out to 200 on each side.
ReplyDeleteTo find the answer to the following question, I used the strategy trial and error. This strategy is when you make repeated attempts until you solve the problem.
ReplyDeleteFirst, I started with 2 numbers. I found it didn't work because the median between the 2 numbers had to be 500. But the closest numbers would be 499+500=999 or 500+501=1001.
Then I moved onto 3 numbers. I found out that first of all, 1000 is not divisible by 3, which automatically eliminates this option.
Moving onto 4 numbers, the median of all these numbers must be 250 because 250 times 4 would be 1000. However, no 4 consecutive numbers add up to 1000. The closest would be 248+249+250+251= 999 or 249+250+251+252=1002.
Now I moved up to 5 numbers. I knew that the median between these 5 numbers would have to be 200. And there was... 198+199+200+201+202=1000.
In conclusion, the least number of consecutive numbers that add up to 1000 is 5.
Grade 8 POTW:
ReplyDeleteHow I did it was to figure out the middle number first, and then subtract to find the smallest number. Since there are 25 numbers and they add up to 1000, the average would be 40. 40 would then be the 13th number. I now subtract 12 from it since the values are consecutive. Because 40 is the 13th number, there would be 12 numbers before it, so I subtract 40 by 12 and get 28.
The smallest value in the 25 consecutive integers is 28.
--ALAN--
I solved the Grade 7 potw with guess and check, or trial and error, and using some background knowledge. I remember from our Data Management unit some knowledge finding the mean and making other sets of numbers with the same mean. For example, the sets of numbers: 25, 25, 25 have a mean of 25. Another set with the same average would be 24, 25, 26. This is because I change one number to be one higher, and the other one lower. This is the strategy I used here. Number of consecutive numbers= How much I divide 1000 by to find the number I have to "build around". I knew 2 would not work because 1000/2=500, and 500+500 does equal 1000, but they are not consecutive. 1000/3=333.333. I knew 3 would not work because 333.333 is a decimal.
ReplyDelete1000/4=250 I knew I could not do four because there is no "center" to work around, as it is an even number.
1000/5=200
I tried using my strategy here:
200+200+200+200+200 to:
198+199+200+201+202 =1000
The minimum amount of consecutive positive integers to sum exactly 1000 is 5.
I found this one using a similar strategy as well!
DeleteThis one was actually pretty easy. At least 2 meant that 1 would not work and I also determined that an even amount of numbers would not work because there is no middle number and 3 would not work as it would force me to decimals. I found that 5 worked and those 5 were: 198,199,200,201,202.
ReplyDeleteThis problem wasn't the hardest in the world, but was a bit challenging. First what I did was trial and error, so 2 numbers, then 3 and on. But we need consecutive numbers, and the ones closest would have been 499+500 or 500+501, but those two don't work. Now we go on to 3, but it isn't divisible, so now 4. But 4 would have as 250*4, so we need a median of 250, but there will be no median. For 5 on the other hand, we are able to get a median of 200, with the numbers 198,199,200,201,202. Therefore, my answer is 198,199,200,201,202.
ReplyDeleteI got 5 consecutive numbers for this POTW. Though I did the work hard copy, i've noticed that many people had similar explanations to the one I had written out. I also went back and finished POTW #10. (I had actually finished it last weekend but forgot to put that up) All math is in my notebook.
ReplyDeleteI was able to get the answer that the smallest positive integer is 28. My guess and check work is on my math book. I also know of a few other formulas but I decided the easiest but more effort rout would be the way to go for this question.
ReplyDeleteI did the work in my math notebook. The answer I got was that you needed a minimum of five numbers.
ReplyDeleteIn order to solve this question, I decided to guess and check. However, in order to guess and check effectively, I needed a number to revolve around. I chose for the number to be the supposed median of the answer (1000/the number of consecutive integers).
ReplyDeleteAnother rule is that the number of integers must be odd and over 1. This is because the consecutive numbers must have the same amount subtracted from the median that is added to the median, in order for the numbers to equal the desired total. For example, 197, 198, 199, 200, 201, 202, and 203 would equal 1400, since the same amount that is added to 200 is the same amount subtracted from 200. However, 199, 200, 201 and 202 would not equal 800, since an equilibrium is not there.
Knowing this, we must find an odd number above 2 with a median. First, I tried the number 3. However, it does not have a median, since 1000 is not divisible by 3. Then, I tried the next odd number, 5. Since this number is odd, above 2, and has a median, 200, you need a minimum of 5 consecutive numbers to reach 1000.
Forgot to write this comment, but i did this on paper. I got 5 consecutive numbers 198,199,200,201,202.
ReplyDeleteI forgot to post this but I did it on paper.
ReplyDeleteI found out the minimum amount of numbers needed to obtain the sun of 1000 was 198,199,200,201 and 202
Im sorry for being late...
ReplyDeleteSo first I used the trial and error method:
After an hour of work, It seemed that it was going to take quite a while if I keep using the same method, but I kept going. First I tested to see what consecutive numbers can add up to near 1000, with the lowest amount used (190), I found my range of numbers at 190-210. Then I tested my method out to see how close to 1000 I was.
After almst 100 tries this is what I got:
198 + 199 + 200 + 201 + 202 = 1000
The smallest amount of positive integers is 5 and they are 198, 199, 200, 201 and 202