POTW #16 Grade 8 Solution:
POTW #16 Grade 7 Solution:
POTW #17 Grade 8:
POTW #17 Grade 7:
Friday, January 20, 2017
POTW #17 - Don't Forget to Try/Review Both Questions and Solutions
Please find the latest POTW below. Remember, BOTH questions provide great practice for a wide variety of math strands/concepts! Be sure to check out Grade 7 & 8 POTW #16 Solutions because they directly involve our next unit in math....Algebra!
POTW #16 Grade 8 Solution:
POTW #16 Grade 7 Solution:
POTW #17 Grade 8:
POTW #17 Grade 7:
POTW #16 Grade 8 Solution:
POTW #16 Grade 7 Solution:
POTW #17 Grade 8:
POTW #17 Grade 7:
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Grade 8 POTW:
ReplyDeleteHow I figured this out was to find out how many multiples of each number there were in 100. Figuring this out would allow me to find out the answer to this question.
Steps to figuring this out:
There is a strategy to figuring out how many multiples of a number there are in 100. The strategy is to use the total number(100 in this case) divided by the multiple. Therefore, to figure out the multiples of 2, I simply use 100/2=50. There are 50 multiples of 2, which can be proved right after double checking.
Because there are also multiples of 3, I have to use 100/3. This gives me 33 with a remainder of 1, but that remainder doesn't really matter. It just means that the remainder isn't a multiple of 3 either. From what I have, I can see that there are 33 multiples of 3 in 100.
So, the answer should be 100-50-33= 17 red cards right? Ehhh.... No. In this question lies a trick, which is that some numbers in 100 are both multiples of 2 and 3! This means that those cards would be flipped twice, and still have the red side face up. The question is, how many numbers in 100 are multiples of 2 and 3? To figure this out, I use the LCM. The LCM of 2 and 3 is 6, so I have to find all the multiples of 6 in 100.
100/6= 16 remainder 4, so there are 16 multiples of 6.
In conclusion, there are 100-50-33+16= 33 numbers that will have the red side facing up.
-Alan-
Grade 7 POTW:
ReplyDeleteThe way I solved it was to find different possibilities to make an even sum out of 4 numbers. There are only 3 ways, 2 even numbers and two odd, 4 odd numbers, and 4 even numbers. However, looking more into the question, I can see that it is impossible to have 4 even numbers because there are only 3, which are 2,4, and 6. So, now I have to make different groups of 4 numbers with the other two ways.
I also have to note that sequences like 4,3,2,1 and 1,2,3,4 are the same, so this question is a combination question. Combination is when order does not matter, and Permutation is when order matters.
There is actually only one selection of 4 odd numbers, because it doesn't matter how you change the places of the 1,3,5, and 7, it would still count as one selection.
On the other hand, the number of different selections for 2 even and 2 odd numbers is 3*2/2*4*3/2=6. 6+1=7, so there are 7 different selections.
The answer should be 7, but I am not fully sure because it is hard to tell whether the question is stating permutation or combination. If it is combination, the answer is 7, but if it is permutation, the answer would be actually 96.
-Alan
I made a miscalculation on the 6, and it is actually 18. So 18+1=19 possibilities.
Delete-Alan
Grade 7 POTW:
ReplyDeleteThe only way to obtain an even sum would be to add 2 odd numbers and 2 even or add all 4 odd numbers.
The only 1 way to add 4 odd numbers is to add 1+3+5+7 which gives a sum of 16 which is even.
There are multiple ways to add 2 odd numbers and 2 even numbers.
1+3+2+4
1+3+2+6
1+3+4+6
1+5+2+4
1+5+2+6
1+5+4+6
1+7+2+4
1+7+2+6
1+7+4+6
3+5+2+4
3+5+2+6
3+5+4+6
3+7+2+4
3+7+2+6
3+7+4+6
5+7+2+4
5+7+2+6
5+7+4+6
There are 18 possible ways 2 odd numbers and 2 even numbers can be added.
18 + 1 = 19.
There are 19 different selections that are possible so the sum of the 4 digits is even.
There are 33 red cards flipped. This is because half of the numbers are multiples of 2 and so half of the cards are red and the other half are yellow. Than because there are 33 multiples of 3 in 100 with 17 being odd. Than because 50-17 is 33.
ReplyDeleteGrade 7 POTW:
ReplyDeleteTo find all the possibilities, there are only 2 ways to finding the sum of 4 numbers is even.
The first way to is to add all the odd numbers together, which is 1+3+5+7. The sum of these numbers is even.
The second way is to add 2 odd numbers and 2 even numbers together. This also gives an even sum.
All the possibilities for adding 2 odd and even numbers are:
1+3+2+4
1+3+2+6
1+3+4+6
1+5+2+4
1+5+2+6
1+5+4+6
1+7+2+4
1+7+2+6
1+7+4+6
3+5+2+4
3+5+2+6
3+5+4+6
3+7+2+4
3+7+2+6
3+7+4+6
5+7+2+4
5+7+2+6
5+7+4+6
There are 18 ways of adding 2 odd numbers and 2 even numbers together. However, we still have to add the first way. 18+1=19
Therefore, there are 19 different selections
For the grade 7 potw, a way to tell if a the number will be an even number is if there are an even number of odd numbers. Doing this there are 19 different solutions.
ReplyDelete1+3+2+4
1+3+2+6
1+3+4+6
1+5+2+4
1+5+2+6
1+5+4+6
1+7+2+4
1+7+2+6
1+7+4+6
3+5+2+4
3+5+2+6
3+5+4+6
3+7+2+4
3+7+2+6
3+7+4+6
5+7+2+4
5+7+2+6
5+7+4+6
1+3+5+7 and since there are only 3 even numbers you can't do 4 even numbers.
Grade 7 POTW:
ReplyDeleteI found out that there are 19 different possibilities of making an even number with 4 numbers from 1-7. The work is in my math notebook
GRADE 7 POTW:
ReplyDeleteThis is Stanley commenting from Derek's Computer. To get an even sum with 4 numbers, there has to be either 0, 2, or 4 odd numbers. If you have 1 or 3 odd numbers, it will be an odd sum. The only problem is that we don't have 4 even numbers, which means there are only 2 ways to get an even sum, and that is having 2 or 4 odd numbers.
1+3+2+4
1+3+4+6
1+3+5+7
1+5+4+2
1+5+6+4
1+5+2+6
Wrote rest on paper. In the end, I ended up with 19 different possibilities.
There are 33 card with the red card facing up in terms of color. This as there are 50 multiples of 2 and 33 multiples of 3. However, there are many common multiples of 2 and 3, so we use the multiples of 6, which are 16 to 100, and we add 16 unique numbers. This gives us an answer of 33 cards.
ReplyDeleteGrade 7 POTW
ReplyDeleteSince there are 4 odd and 3 even numbers, the only way to get an even result is by adding 2 odd and 2 even numbers together or adding the 4 odd numbers together. I decided to make a list of all the possibilities that fulfill the conditions above.
1+2+3+4
1+2+3+6
1+2+5+6
1+2+5+4
1+2+7+6
1+2+7+4
1+4+3+6
1+4+7+6
3+2+5+4
3+2+5+6
3+2+7+4
3+2+7+6
3+7+4+6
3+5+4+6
5+2+7+4
5+4+7+6
5+7+2+6
1+3+5+7
Therefore, there are 19 different solutions that will result in the sum of the 4 integers being even.
To complete this question I basically divided 100 by 2 to find out how many multiples of 2 are in there which is 50. Then after that I imagine 50 cards flipped over. After that I think of all the multiples of 3 (33). Then I picture that many cards flipped over. Thus there are 83 cards currently flipped over. The catch is that I already flipped over 50 cards and 3 and 2 have cross over multiples. So I found the multiples of 6 also (16). Then I flipped over those cards back from the 83 thus making the yellow cards 67. After that I subtracted 65 from 100 to get 33 since we are looking for the red cards. There are 33 red cards.
ReplyDeleteGrade 8 POTW:
ReplyDeleteHow I solved this by first splitting the the question is half and finding how many times the cards are flipped.
The first time, 50 cards were flipped. I can tell ths because there were 100 cards, and we were supposed to flip every multiple of 2, or every secong number, so 100/2 = 50.
The second time, 33 cards were flipped. can say this because every multiple of 3 or every 3rd number is flipped. Therefore 100/3 = 33.33, but we cant flip half cards so we only flip 33.
But, some cards that were flipped were both multiples of 2, and 3. So I found the LCM of 2 and 3 which was 6.
So I found the multiples of 6 from 1 - 100. Therefore 100/6 which is 16 with a remainder of 4. So 16 cards were flipped twice
Therefore at the end the equation was 100-50-33+16 which is 33.
At the end, there are 33 cards that have the red side facing upwards.
-Khush :)
Here is how I solved the Grade 8 POTW:
ReplyDeleteso first there are 50 multiples of 2 in 100 ( because 100/2 is 50)
there are 33 multiples of 3 in 100 ( because 100/3 is 33)
and 6 is the lowest multiple of three and two, so there are 16 multiples of 6 in 100 ( 100/6 is 16)
So take away 16 from 33 because that 16 numbers is already in the multiples of 2.
Meaning 50 + 17 is 67 so those are the cards that have the yellow side up.
100 - 67 = 33
So 33 cards have the red side faced up.
Here is how I went about solving the Grade 7 POTW:
ReplyDeleteI started out by realizing that to get a even sum, you need to have either all odd numbers, all even numbers, or 3 odds and an even. Using these rules I replaced the same spot that previously had an odd number, with another odd number. I got the order (vertically) by alternating from one odd number to the next and switching up the spots. I would say my main strategy was to guess and check.
1) 1+3+5+7
2) 1+3+2+4
3) 1+3+2+6
4) 1+3+4+6
5) 1+5+2+4
6) 1+5+2+6
7) 1+5+4+6
8) 1+7+2+4
9) 1+7+2+6
10) 1+7+4+6
11) 3+5+2+4
12) 3+5+2+6
13) 3+5+4+6
14) 3+7+2+4
15) 3+7+2+6
16) 3+7+4+6
17) 5+7+2+4
18) 5+7+2+6
19) 5+7+4+6
Therefore, there are 19 possible combinations.
Grade 7 POTW
ReplyDeleteI solved this because I know that to get an even number with 4 integers, it has to be 4 odds, 4 evens, or 2 evens and 2 odds. But since there are only 3 evens, there has to be an odd but that odd is going to make the sum an odd number so 4 evens won't work. With the process of elimination...
1+3+5+7
1+3+2+4
1+3+2+6
1+3+4+6
1+5+2+4
1+5+2+6
1+5+4+6
1+7+2+4
1+7+2+6
1+7+4+6
3+5+2+4
3+5+2+6
3+5+4+6
3+7+2+4
3+7+2+6
3+7+4+6
5+7+2+4
5+7+2+6
5+7+4+
I ended up with 19 different combinations.
Odd Even
ReplyDelete1+2 1+7
2+3 2+6
3+4 3+5
4+5 6+4
5+6 1+3
6+7 5+7
1+5
1+7
Using this chart, there are a total of 19 combinations with each side combining 2 of it's own side.
Sorry for delayed answer.