Grade 7 POTW #13 Solution:
Grade 8 POTW #13 Solution:
Fiona 7/8 Solution #1:
This question seems more difficult than it is. All we know is that there is an 8 x 8 square
and the sum of the four corners of 1646. We can use algebra to find the value of the
top left corner and use that to find the value of the bottom right corner. The top left
corner has to be a variable, let’s say x. The expression for the top right corner
would be x + 7 since you need 8 numbers inclusive (including the first and the last).
The expression for the bottom left would be x + (7 x 24) or x + 168, since from the
top row to the second row (same column), there is 24 that is added on. Since you
are moving 7 rows down (for the same reason you added 7 for the top right), you multiply
the two to get
and the sum of the four corners of 1646. We can use algebra to find the value of the
top left corner and use that to find the value of the bottom right corner. The top left
corner has to be a variable, let’s say x. The expression for the top right corner
would be x + 7 since you need 8 numbers inclusive (including the first and the last).
The expression for the bottom left would be x + (7 x 24) or x + 168, since from the
top row to the second row (same column), there is 24 that is added on. Since you
are moving 7 rows down (for the same reason you added 7 for the top right), you multiply
the two to get
x + 168. The bottom right uses the expression for the bottom left, and adds an
extra 7, making the expression x + 175. If I put all of these expressions into an
equation, we can find the answer.
extra 7, making the expression x + 175. If I put all of these expressions into an
equation, we can find the answer.
Let x be the value of the top left corner.
x + (x + 7) + (x + 168) + (x + 175) = 1646
4x + 350 = 1646
4x = 1296
x = 324
Knowing this, we can substitute x with the equation that we put in for the
bottom right corner.
bottom right corner.
Bottom right:
x + 175
= (324) + 175
= 499
Therefore, the bottom right corner is 499.
Grade 7 POTW #14 Question:
Grade 8 POTW #14 Question:
Grade 7/8 POTW Fiona Question #2:
Question: In the diagram, BD is perpendicular to BC and to AD. If AB = 52,
and AD = 48, what is the length of DC?
(if the image does not show up on your screen open this
https://docs.google.com/document/d/1FUNukUoynkVY0RRCGMDfzeaVZmhTf6Tvxwbfujh6qjc/edit )
https://docs.google.com/document/d/1FUNukUoynkVY0RRCGMDfzeaVZmhTf6Tvxwbfujh6qjc/edit )
Grade 8 POTW:
ReplyDeleteInfo:
- One person lied in both
- One person told the truth in both
- The other 2 lied once and truthed once (yes I know “truthed” is not a word)
Andy: “Barb lied once” “Dana lied twice”
Barb: “I never lie” “Andy never lied”
Carl: “Dana lied twice” “Barb never lied”
Dana “Andy lied twice” “I never lie”
Since this isn’t as straightforward as most math is, we need to consider the different scenarios that could happen and see if everything lines up. Once we find one part of the solution, most of the solution should fall into place.
Let’s say we assume that Andy lied twice.
So if what he said about Barb lying once is a lie, we know that Barb either lied twice or never lied. But since Andy is the one that “lied twice”, Barb has to be telling the truth. But this would contradict Barb saying that Andy never lied. Therefore, Andy cannot be the one to lie twice.
Let’s assume that Barb lied twice. Then the Andy never lied would be a lie, leaving Andy with lying once or lying twice. Andy could not have lied twice because Barb already “lied twice”. But Andy lying once is also a lie because Barb lied once/twice. “Dana lied twice” is also a lie because Barb lied twice in this scenario. Barb could not have lied twice.
Let’s assume that Carl lied twice. This means that Barb lied either once or twice and Dana lied once or not at all. Barb could not have lied twice because Carl did. But if Andy was telling the truth, it would contradict the “Dana lied twice. Carl did not lie twice.
This only leaves Dana to lie twice. Now we need to find who told the truth for both.
Automatically, we can cross out Barb because if they were telling the truth, then both them and Andy would have never lied. This cannot be the case since only one person didn’t lie. Barb did not tell the truth for both.
For Carl, it is similar. The statement about Dana is true, but the statement that Barb never lied is false because that would mean that Barb and Carl both did not lie, which cannot happen since only one person can “not lie”.
The only person left is Andy.
Therefor, Dana lied twice, Andy didn’t lie and Barb and Carl each lied once.
Yay my POTW is up :)
ReplyDeleteI'll try to find more possible POTWs for the Grade 7/8s to do. :))
Happy to contribute!
C:
So for the truth and lying question here is what I did:
ReplyDeleteIf andy told the truth twice then:
Barb's firwst statement would have to be false since there is only one person who never lied, while his second statement which would be a truth, would be that Andy never lied, which I am hypothesizing right now that he didn't. Furthermore, Andy's first statement would support that Barb only lied once.
Carl's first statement would be impossible to find out unless we look at Dana's statement, while his second statement had to be false, since Andy was the one who never lied.
Now, Dana's first statement would have to be false, and her second statement would also be false, since we know Andy was the person who didnt lie, thus, we I can see the Carl's first statement was true, and that he and Barb was the person who lied once and told the truth once. Andy was the person who only told the truth, and Dana always lied (darn you dana) (no offense).
Barb can't be the person who always told the truth, since he already states that two people never lied, while the question conflicts with that, stating only one person always told the truth.
Carl can't be the person always telling the truth since he also says that Barb never lied, which already makes two people never lie before.
If Dana always told the truth;
Her first statement says that both of Andy's statements are lies, and Andy's first statement said that Dana lied twice, which is false, but he also says that Barb lied once, which had to be false because of Dana's accusation. Thus, Barb had to be either the person who always lied or the person who always told the truth. He can't be both since there is only one person being both; Dana is the person telling the truth (hypothetically in this scenario), and Andy is the one always lying, meaning that Dana can't be the person always lying.
This means the person who was telling the truth was Andy, the person always lying was Dana, and the people lying once and telling the truth once was Barb and Carl
The "Fiona Question" for this week was actually kind of easy. First, I found the length of BD using the Pythagorean theorem. I did 52 squared and 48 squared, which turned out to be 2704 and 2304 respectively. Since 52 is the length of the diagonal, I had to subtract 2304 from 2704, which gave me 400 as the square of the length of BD, instinctively, I knew that the length was 20.
ReplyDeleteNow, I use the Pythagorean theorem again to find the length of DC.
The square of 20 is 400 and the square of 21 is 441. The total is 841, meaning the square root of 841 is the length of DC. At the beginning, I thought the square root of 841 would be an irrational number, but after doing some calculations, I found that 29 was the perfect square root of 841.
Thus, the length of DC is 29 units.
We know that for a number to be divisible by 5, it needs to have either the number 5 or 0 as the last number, and because it's 7 digits, the number cannot be 0. Now we have the number format 5xyzyx5. We know that numbers divisible by 15 include 15, 30, 45, 60, 75, 90..., However, using the digit root is a more effective method. Let's set x and y to 9 and see what happens:
ReplyDelete|599z995| = 36.
Now, all we need to do is find the middle number via trial and error, starting from 9.
5999995/5 = not a whole integer
5998995/5 = a whole integer, 1199799.
Therefore, the largest palindromic number divisible by 15 is 5998995.
To find the smallest, we substitute 0 as x and y, and try to find the smallest amount.
z=1 does not work
z=2 works.
Therefore, the smallest palindromic number divisible by 15 is 5002005.
Thanks for tuning in!
(hope I didn't type any palindromes wrong)
Tin's POTW for the Grade 7 Question:
ReplyDeleteLargest 7-digit palindrome divisible by 15:
So first, I knew that a number divisible by 15 is basically a number that is divisible by 5 and 3. All numbers divisible by 5 end with either a 5 or a 0, and since if the end is 0, then the first number would be 0 because that is how palindromes work, but then it wouldn't really be a 7-digit with 0 at the beginning, also, 5 is larger than 0. So now, the largest possible palindrome divisible by 5 would be 5999995, since the first number had to be 5, because the end was five, and other integers I just used 9's since they were the largest single digit. However, that number is not divisible by 3, and I used the trick where all numbers whose digits add up to a number divisible by 3 IS divisible by 3. I added up the digits of 5999995, which gave me 55. The closest numbers divisible by 3 were either 57 or 54. I couldn't add 2 more to any of the integers since 9 already is the largest, so I subtracted 1 from the middle, since the middle (4th digit) is the only digit where its change would not affect any other digits. So the largest number for a 7-digit palindrome divisible by 15 is 5998995, ad to check that, I divided it by 5, which gave me a whole number, 399933. The smallest number also had to end with either 0 or 5, since it had to be divisible by 15. If it ended with 0, then the beginning would be a 0, which wouldn't make sense as a number, that meant it had to start and end with 5. For the other digits, I naturally chose 0 for all of them since 0 is the smallest possible digit. This gave me 5000005, which was not divisible by 5. I again used the strategy to check if a number is divisible by 3 to help me. The sum of the digits of the number was 10, meaning I either had to subtracted 1 so it becomes 9, or add 2 so it becomes 12. I couldn't subtract 1 since there were no digits suitable to do that, meaning I had to add 2. This meant I replaced the middle 0 with a 2, which made the number (5002005) divisible by 15.
So the conclusion:
The largest 7-digit palindrome number divisible by 15 is 5998995, while the smallest is 5002005.
The divisibility rule for 15 is that if the number is divisible by 3 and 5.
ReplyDeleteThe rule for 3 is if the digits add up to a number divisible by 3 and 5 is if the 1's digit is 0 or 5.
This means that the number has to start with a 5 because it has to end in a 5.
5xyyx5 is the template that we need to use
because it also has to be divisible by 3, x+y+y+x+10=something divisible by 3
the numbers divisible by 3 after 10 are: 12,15,18,21,24
to find the smallest you make either x or y 0 and the other one 1.
That makes 501105 or 510015.
To find the largest one you find the biggest numbers
597795 is one of the numbers. The sum= to 42
I did 6 digits the first time....
ReplyDeleteSo lemme redo this
the starting number has to equal 5 because the end has to equal 5
5xyzyx5 or 5yxzxy5
Smallest is x or y =1 and the others equal 0
510015 or 501015
Biggest is
59995 doesn't work so then subtract 1 by subtracted 1 from z
59895 works because it ends with 5 and 5+9+8+9+5=36 which is divisible by 3
Grade 8 POTW
ReplyDelete- One person lied in both
- One person told the truth in both
- Two people told the truth and lied once
To solve this, I am just going to use facts and possibilities and see if they will work. I'm also going to go straight into it.
Suppose Andy tells the truth twice. This means Barb lied once and Dana lied twice. Barb said she never lies and Andy never lies. To make this true, the "I never lie" statement must be a lie. If Andy doesn't lie, it also means that Dana never told the truth. Dana said Andy lied twice and she doesn't lie, and if both are lies, that makes sense. So far, it makes sense. The only possible outcome now is if Carl lies once. Carl said Dana lied twice, and Barb never lied. We know Dana did lie twice and Barb DID lie meaning he is did lie once. Surprisingly, we got the answer in the first attempt.
Therefore, Andy never lied, Barb lied exactly once, Carl lied exactly once, and Dana lied twice.
Grade 7 POTW
ReplyDeleteI will just go straight into it. We have to find the largest seven digit palindrome divisible by 15 and the smallest divisible by 15. In order for something to be divisible by 15, it has to be divisible by 3 and 5. To be divisible by 5, the number has to end with a 5 or 0. In this case, 0 can't be the end as it would be the beginning and 0 can't begin a number (unless in the case if it is 0 or a decimal). Anyways, the value so far is: 5#####5 for both. For the largest number we can start by maximizing the value by inserting only 9s. In other words, we'll start with 5999995. For it to be divisible by 3, the digit sum must be divisible by 3. The digit sum for this is 55 which isn't divisible by 3 (to keep the whole value). The next lowest digit sum we can aim for that is divisible by 3 is 54 as 57 requires increasing the value which isn't possible. We have to change one digit to one less or other digits to much less. We are trying to maximize the value so the only thing to do is lower the middle digit down one (9 - 1 = 8). 5998995. Moving on to the lowest number, we can just do the opposite of before and minimize the number to begin with. In other words, we'll start with 5000005. The digit sum is 10. We can't decrease it, so we'll have to increase it to 12. 12 - 10 = 2 so we have to increase 2 digits by 1 or 1 digit by 2. If we increase the second last digit by 1, the second digit will be increased by 1, but it wouldn't be the lowest. Moving one closer to the middle, the same result occurs as it still can be lowered. To minimize the number, we have to increase the middle digit by 2. 0 + 2 = 2. 5002005.
Therefore, the greatest 7 digit palindrome divisible by 15 is 5998995 and the smallest 7 digit palindrome divisible by 15 is 5002005.
Grade 7/8 POTW
ReplyDeleteI will go straight into the problem. The goal is to figure out the length of DC. To do this, we can firstly, find the value of BD using the Pythagorean theorem (a^2 + b^2 = c^2). We can call a 48 units and c (hypotenuse) 52 units. To solve for b, we have to change the equation slightly (c^2 - a^2 = b^2). 52^2 - 48^2 = 2704 - 2304 = 400. Square root of 400 = b = bd = 20 units. Now using even more Pythagorean theorem, 20^2 + 21^2 = c^2 = 400 + 441 = 841. Square root of 841 = cd = 29.
The length of cd is 29 units.
Grade 7 POTW:
ReplyDeleteInfo:
- 7 digits
- Palindrome divisible by 15
Basically, we are looking for the smallest and largest numbers that are divisible by 3 and 5 both, and are the the format of abcdcba. Using divisibility rules, we can rule out (ha, get it?) the possibility of the number ending in 0 because it would have to also start with 0, but that 0 would be useless and would cancel out, making it a 6 digit number (what a run-on sentence). a must be 5.
Number: 5bcdcb5
To find the smallest possible number, we need to make b and c the same number, 0.
Number: 500d005
Now, we need to find the smallest number thatch fit into d so that the number is divisible by 3.
5+5+0+0+0+0+d= 10+d
The smallest possible number for d is 2 (10+2=12. 12/3=integer).
Number:5002005 (insert checkmark here)
Now, we need to find the largest number.
Number:599d995
5+5+9+9+9+9+d = 46+d
d can be 8 because that would give 54 and 54 is divisible by 3.
Number: 5998995
The smallest number is 5002005 and the largest number is 5998995.
Grade 7 POTW
ReplyDeleteIn order to find the smallest and largest 7 digit palindromic number, I have to use ABCDCBA to get my answer. A is required to be 5 because we are looking for numbers divisible of 3 and 5(3x5=15, 5x3=15) and A cannot be 0 because a palindrome with 0 as it's first number would mean that the number is 6 digits and not 7. The new number is 5BCDCB5. Since 0 is the smallest number from 0 to 10, I will use it to make the next 2 variables and the second and third last variables 0. The new number is 500D005. Since 15 is divisible by 3, I have to add up all of the known numbers (5+0+0+0+0+5+D), and then I have make D a number to get the sum to be 12 which is the next number divisible by 3 (12/3=4). The smallest number is 5002005. Now to find the largest number, I have to change the 0s to 9s and then follow the same process to find D. 5+9+9+9+9+5=46 + 8 which is the highest single digit number that makes the sum of all of the digits 54 which is divisible by 3. 54/3=18.
The smallest and largest 7 digit palindromic numbers divisible by 15 are 5002005 and 5998995.
Starting off, here is what I know: I know that if a number is divisible by 15, it must end in a 5 or 0. However, because this is a palindrome, it would start with these. But in order to be 7 digits, the number cannot start with 0. This means they will both start with 5. From their, I just used trial and error and dividing the possibilities by 15 until I got my answer. I found the least in value one to be: 5002005 and the greatest to be: 5998995.
ReplyDeleteTo make this easier we will write this as a variable:
ReplyDeletea b c d c b a
Now since is must be divisible by 5 than the last number can only be 5 or 0. Because the first digit cannot be 0 than a equals 5.
5 b c d c b 5
Now to find the smallest we must make it divisible by 3. The digits must add up to a multiple of 3.
5002005 should do the trick: 5+0+0+2+0+0+5 = 12 = multiple of 3
To find the largest I used trial and error to find the largest number that was divisible by 3.
THE SMALLEST:5002005
THE LARGEST:5998995
This comment has been removed by the author.
ReplyDeletefor it to be the largest palindrome divisible by 15 it has to end in 5 meaning the first digit has to be 5.
ReplyDeletenow in the middle you try to add as much as of the highest digit possible.
5999995
But the digits add up to= 55
than subtract one from it to get the 54 which is divisible by 3.
so change the middle digit to 8
there for my largest number is 5998995
For my smallest I found the smallest number over 10 divisable by 3 that was 12.
So 5+5+2=12
5002005 is the smallest number
Therefore the largest number is 5998995 and the smallest is 5002005
POTW:
ReplyDelete- The palindrome needs to be divisible by 15
- To be divisible by 15, it needs to be divisible by 5 and 3
- 5 cannot be divided by zero
- so the digit can't start with zero.
From there, I found some combinations, using trial and error, the least being 5002005, and greatest 5998995.