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Grade 7 POTW I answered this question was educated guess and check. I started off by making C=3 and B=5. I realized that it was too low because each row had to be 21 and the diagonal DCB was only equal to 9. Then, I kept raising C and B until I got to C=9 and B=11. I know this is correct because each row had to be 27 and all of them were 27, even DCB(D=7 + C=9 + B=11 = 27). Now, I can add A and E. 15 + 17=32. The answer to this question is 32.
First we need to find out how many possible three-digit numbers can be formed. There are 7 total numbers so… 7 possibilities for the first digit times 6 possibilities for the second digit times 5 possibilities for the third digit equals a total of 210. Each digit has an equal chance of being a units, tens or hundreds digit. So… 210/7 = 30 times the digit appears per place value.
For this question, we need to use the numbers that have a repeating digit for all the place values. These numbers are 111,222,333,444,555,666,777. The highest possible number to make would be 777, and the lowest possible number would be 111. If we continue the pattern, we get the rest of the numbers. If we add up these numbers we would get to 3108. This would only account for some of the sums. But we still have that 30 from the beginning, so we can multiply them together to get 93240.
Grade 8 POTW Okay, so Matt adds every 3 digit combination of numbers from 1-7 and gets a sum. The goal is to find an easy way to get this sum. Firstly, we know there are exactly 7 x 6 x 5 = 210 3 digit numbers that can be made. The seven represents the 7 possible first digits, the 6 represents the remaining possible second digits, and the 5 represents the remaining possible third digits. In each of these 210 combinations, each digit appears an equal number of times per hundreds, tens, and ones digit. Therefore, for the 210 numbers, each digit appears 210/7 = 30 times for each of these digits. We can now use this to find out the sum. Each digit appears exactly 30 times each for the ones digits so: 30(1 + 2 + 3 + 4 + 5 + 6 + 7) = 30(28) = 840 Each digit appears exactly 30 times each for the tens digits so: 30(10 + 20 + 30 + 40 + 50 + 60 + 70) = 30(280) = 8400 Each digit appears exactly 30 times each for the hundreds digits so: 30(100 + 200 + 300 + 400 + 500 + 600 + 700) = 30(2800) = 84000 Now when we add these values up, we get: 30(1 + 2 + 3 + 4 + 5 + 6 + 7) = 30(28) + 30(10 + 20 + 30 + 40 + 50 + 60 + 70) + 30(100 + 200 + 300 + 400 + 500 + 600 + 700) = 30(111 + 222 + 333 + 444 + 555 + 666 + 777) = 30(3108) = 840 + 8400 + 84000 = 93240 Therefore, an easier way of adding these all up is to add up the actual digit values for however many times it appears (30 times each in this question), and then add up the values at the end. Using this method, we get the sum of 93240.
Grade 7 POTW The question states that the first 9 positive odd integers are placed in this table. They are 1, 3, 5, 7, 9, 11, 13, 15, and 17. We can find the total sum of the entire table by adding 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 81. We can then divide this by 3 to get 81/3 = 27 = Sum of each row, column and diagonal. Using this, we can easily solve for each variable: B: 13 + 3 + B = 27 16 + B = 27 B = 11 A: 1 + A + B = 27 1 + A + 11 = 27 12 + A = 27 A = 15 C: 5 + 13 + C = 27 18 + C = 27 C = 9 E: 1 + C + E = 27 1 + 9 + E = 27 10 + E = 27 E = 17 (I actually don’t need to find D as a) I already have A and E values, and b) it is the last remaining value, 7) Now we just need to add up A and E: A + E = A + E 15 + 17 = A + E 32 = A + E Therefore, the sum of A and E is 32.
Grade 7 POTW The question states that the first 9 positive odd integers are placed in this table. They are 1, 3, 5, 7, 9, 11, 13, 15, and 17. We can find the total sum of the entire table by adding 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 81. We can then divide this by 3 to get 81/3 = 27 = Sum of each row, column and diagonal. Using this, we can easily solve for each variable: B: 13 + 3 + B = 27 16 + B = 27 B = 11 A: 1 + A + B = 27 1 + A + 11 = 27 12 + A = 27 A = 15 C: 5 + 13 + C = 27 18 + C = 27 C = 9 E: 1 + C + E = 27 1 + 9 + E = 27 10 + E = 27 E = 17 (I actually don’t need to find D as a) I already have A and E values, and b) it is the last remaining value, 7) Now we just need to add up A and E: A + E = A + E 15 + 17 = A + E 32 = A + E Therefore, the sum of A and E is 32.
Info: - First 9 positive odd integers - 3x3 - Same sum of each row, column and diagonal.
The first 9 positive odd integers are as follows: 1,3,5,7,9,11,13,15 and 17. To find the sum of each row, column and diagonal, we need to add the numbers up and divide them by 3 to find the total sum of each row, column and diagonal (3 represents these three…. things). Sum = 81. 81/3 = 27. 27 = sum of each row, column and diagonal.
Now, we can solve for each variable, starting with B: 27 - 3 - 16 = 11
A = 27 - 1 - 11 A = 15
C = 27 - 5 - 13 C = 9
E = 27 - 9 - 1 E = 17
A + E = 15+ 17ds A + E = 32
Joke: I can't help but get the shivers when I'm around THREE. She seems very ODD
Grade 7 POTW
ReplyDeleteI answered this question was educated guess and check. I started off by making C=3 and B=5. I realized that it was too low because each row had to be 21 and the diagonal DCB was only equal to 9. Then, I kept raising C and B until I got to C=9 and B=11. I know this is correct because each row had to be 27 and all of them were 27, even DCB(D=7 + C=9 + B=11 = 27). Now, I can add A and E. 15 + 17=32. The answer to this question is 32.
Grade 8 POTW:
ReplyDeleteFirst we need to find out how many possible three-digit numbers can be formed. There are 7 total numbers so…
7 possibilities for the first digit times 6 possibilities for the second digit times 5 possibilities for the third digit equals a total of 210. Each digit has an equal chance of being a units, tens or hundreds digit. So… 210/7 = 30 times the digit appears per place value.
For this question, we need to use the numbers that have a repeating digit for all the place values. These numbers are 111,222,333,444,555,666,777. The highest possible number to make would be 777, and the lowest possible number would be 111. If we continue the pattern, we get the rest of the numbers. If we add up these numbers we would get to 3108. This would only account for some of the sums. But we still have that 30 from the beginning, so we can multiply them together to get 93240.
That truly is sum-thing. (wow i’m so funny)
Punny!
DeleteGrade 8 POTW
ReplyDeleteOkay, so Matt adds every 3 digit combination of numbers from 1-7 and gets a sum. The goal is to find an easy way to get this sum. Firstly, we know there are exactly 7 x 6 x 5 = 210 3 digit numbers that can be made. The seven represents the 7 possible first digits, the 6 represents the remaining possible second digits, and the 5 represents the remaining possible third digits. In each of these 210 combinations, each digit appears an equal number of times per hundreds, tens, and ones digit. Therefore, for the 210 numbers, each digit appears 210/7 = 30 times for each of these digits. We can now use this to find out the sum. Each digit appears exactly 30 times each for the ones digits so:
30(1 + 2 + 3 + 4 + 5 + 6 + 7) = 30(28) = 840
Each digit appears exactly 30 times each for the tens digits so:
30(10 + 20 + 30 + 40 + 50 + 60 + 70) = 30(280) = 8400
Each digit appears exactly 30 times each for the hundreds digits so:
30(100 + 200 + 300 + 400 + 500 + 600 + 700) = 30(2800) = 84000
Now when we add these values up, we get:
30(1 + 2 + 3 + 4 + 5 + 6 + 7) = 30(28) + 30(10 + 20 + 30 + 40 + 50 + 60 + 70) + 30(100 + 200 + 300 + 400 + 500 + 600 + 700) = 30(111 + 222 + 333 + 444 + 555 + 666 + 777) = 30(3108) = 840 + 8400 + 84000 = 93240
Therefore, an easier way of adding these all up is to add up the actual digit values for however many times it appears (30 times each in this question), and then add up the values at the end. Using this method, we get the sum of 93240.
Grade 7 POTW
ReplyDeleteThe question states that the first 9 positive odd integers are placed in this table. They are 1, 3, 5, 7, 9, 11, 13, 15, and 17. We can find the total sum of the entire table by adding 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 81. We can then divide this by 3 to get 81/3 = 27 = Sum of each row, column and diagonal. Using this, we can easily solve for each variable:
B:
13 + 3 + B = 27
16 + B = 27
B = 11
A:
1 + A + B = 27
1 + A + 11 = 27
12 + A = 27
A = 15
C:
5 + 13 + C = 27
18 + C = 27
C = 9
E:
1 + C + E = 27
1 + 9 + E = 27
10 + E = 27
E = 17
(I actually don’t need to find D as a) I already have A and E values, and b) it is the last remaining value, 7)
Now we just need to add up A and E:
A + E = A + E
15 + 17 = A + E
32 = A + E
Therefore, the sum of A and E is 32.
Grade 7 POTW
ReplyDeleteThe question states that the first 9 positive odd integers are placed in this table. They are 1, 3, 5, 7, 9, 11, 13, 15, and 17. We can find the total sum of the entire table by adding 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 81. We can then divide this by 3 to get 81/3 = 27 = Sum of each row, column and diagonal. Using this, we can easily solve for each variable:
B:
13 + 3 + B = 27
16 + B = 27
B = 11
A:
1 + A + B = 27
1 + A + 11 = 27
12 + A = 27
A = 15
C:
5 + 13 + C = 27
18 + C = 27
C = 9
E:
1 + C + E = 27
1 + 9 + E = 27
10 + E = 27
E = 17
(I actually don’t need to find D as a) I already have A and E values, and b) it is the last remaining value, 7)
Now we just need to add up A and E:
A + E = A + E
15 + 17 = A + E
32 = A + E
Therefore, the sum of A and E is 32.
Grade 7 POTW:
ReplyDeleteInfo:
- First 9 positive odd integers
- 3x3
- Same sum of each row, column and diagonal.
The first 9 positive odd integers are as follows: 1,3,5,7,9,11,13,15 and 17. To find the sum of each row, column and diagonal, we need to add the numbers up and divide them by 3 to find the total sum of each row, column and diagonal (3 represents these three…. things). Sum = 81. 81/3 = 27. 27 = sum of each row, column and diagonal.
Now, we can solve for each variable, starting with B:
27 - 3 - 16 = 11
A = 27 - 1 - 11
A = 15
C = 27 - 5 - 13
C = 9
E = 27 - 9 - 1
E = 17
A + E = 15+ 17ds
A + E = 32
Joke: I can't help but get the shivers when I'm around THREE. She seems very ODD