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Grade 7 POTW: First I need to find common multiples with 21 and 35 if I want to find a number that has divisors of 21 and 35. The LCM is 105 (I prime factorize the two numbers and multiply the prime factors together to get 105) Now, I need to factor 105 to see if it has 8 divisors. 105= 1,3,5,7,15,21,35,105 Oh would you look at that, there are 8 positive factors!!!! Oh wow!!! The answer is 105.
Sorry, I forgot to do this POTW so I will do it now (better then never:D)
Info: - Neither team scored three goals in a row - There were six goals scored in the match. - Find all possible scores - Find how many the Beaver Rangers won
Beaver Rangers: 1 Fox Raiders: 2
Possible Combinations: 1, 1, 2, 1, 1, 2 B 2, 1, 1, 2, 1, 2 Tie 1, 1, 2, 1, 2, 2 Tie 2, 1, 1, 2, 2, 1 Tie 2, 2, 1, 2, 1, 1 Tie 1, 1, 2, 2, 1, 1 B 1, 2, 1, 1, 2, 2 Tie 1, 1, 2, 1, 2, 1 B 1, 1, 2, 2, 1, 2 Tie 1, 2, 1, 1, 2, 1 B 1, 2, 1, 2, 1, 2 Tie 1, 2, 1, 2, 2, 1 Tie 2, 2, 1, 2, 2, 1 F 2, 2, 1, 1, 2, 1 Tie 2, 2, 1, 1, 2, 2 F 1, 2, 2, 1, 1, 2 Tie 1, 2, 2, 1, 2, 1 Tie 2, 1, 2, 1, 1, 2 Tie 2, 1, 2, 1, 2, 1 Tie 1, 2, 2, 1, 2, 2 F 2, 1, 2, 1, 2, 2 F 2, 1, 2, 2, 1, 1 Tie 2, 1, 2, 2, 1, 2 F 2, 1, 1, 2, 1, 1 B 2, 2, 1, 2, 1, 2 F 1, 2, 1, 2, 1, 1 B There are 26 combinations in which 6 of them are won by the Beaver Rangers.
The mystery number is 105. At first, I multiplied 21 and 35 together to see how many factors would arise from its product. The answer to that was 12. These factors: 1, 3, 5, 7, 15, 21, 35, 49, 105, 147, 245, 735, are all divisors of 735 (21 * 35). Since that didn't work, I decided to find the LCM (lowest common multiple) of the two numbers. Jackpot!!! 105 (the LCM) had exactly 8 positive factors: 1, 3, 5, 7, 15, 21, 35, 105. Seeing as it matches the criteria, I can now be sure that the "integer" is 105.
Grade 7 POTW First, I'll list the facts - Integer has exactly 8 divisors - 2 divisors include 21 and 35 Okay. Knowing that 21 and 35 are a divisor means that a common multiple between the two numbers would result in the number. There are only 8 divisors though so the number won't be too high since the higher you go numerically would most likely have more factors. Also, I know that I only need to find 4 factors because I already somewhat know 4. There is 1, 21, 35, and the number itself. I'll start with the lowest common multiple. 21, 42, 63, 84, 105, 126, 147... 35, 70, 105, 140... So far there is 1 common multiple. 105 Factors of 105: 1, 3, 5, 7, 15, 21, 35, 105 105 is the number.
Grade 8 POTW First, I'll list the facts. - Digit product of 2000 - Lowest integer possible This is going to need some more facts too. 1. Not a 1 or 2 digit because for example, 99 (the highest) has a digit product of 99. 2. No 0s. 0s result in 0 as the product. 3. I need to find all factors of 2000 because no matter what, a number must be a factor of 2000 to have a digit product of 2000. 4. No 2 digit factors of 2000. They will be counted as individual factors 1, 2, 3, 5, 8 are the only digits that are usable (I'm pretty sure), however, others can be used because of multiples and different ways to multiply, and so on, so I'll use them all. Let's use those to create the integer, but first, I need a way to multiply them to get 2000. I'm going to see powers of each digit to make it easier. 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048 3, 9, 27, 81, 243, 729, 2187 4, 16, 64, 256, 1024, 4096 5, 125, 625, 3125 6, 36, 216, 1296, 7776 7, 49, 343, 2401 8, 64, 512, 4096 9, 81, 729, 6561 Now, I just need to make 2000. I didn't know why I didn't think of this earlier, but I need to do prime factorization. 2, 2, 2, 2, 5, 5, 5 Now, I will do something that the question didn't say not to do. It's a positive integer. Here is the answer I believe. 1.111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111... You get the point it continues on infinitely and then the last digits are this: 1111111111111111111111111111111111111111111111111111111111111111111111...2222555. The digit product is still 2000! The final lowest integer is 1.111111...2222555.
Another mistake. I didn't really "re-do" my answer properly. The old one would've been right if decimals were allowed, but I can't just get rid of the other digits. 2222555 can easily become 25558. Final re-answer is 25558 (unless I mess up again).
21 35 LCM: 105 105/21= 5 105/35= 3 1,3,5,7,15,21,35,105 There are 8 divisors in 105. The two divisors 21 and 35 are in the factors of 105. I got this by finding the LCM (Lowest Common Multiple) of 21 and 35. The LCM is 105. I then found the factors of 105 which are 1, 3, 5, 7, 15, 21, 35, and 105. There are 8 factors of 105, making it the answer.
I did a similar proccess of Edwyn finding a the smallest common multiple between teh to intergers displayed. I got 105. This even number has 8 divisors and we have listed 2. The other divisors are 5, 1, 15, 35, 105 and 3
First I will find the LCM of 21 and 35. 21: 21, 42, 63, 84, 105 35: 35. 70, 105 The LCM is 105. The factors/divisors of 105 are 1, 3, 5, 7, 15, 21, 35 and 105. The integer is 105.
Grade 7 POTW: First of all, I would find the least common multiple, but I would rather do it an easier way than guess and check. I prime factorize both numbers, 21 and 35 since the mystery number is divisible by both of them. This is simplified to 3*7 and 7*5. I can get rid of one of the 7's because both numbers have it in common so I only need one. This means that I am left with 3*5*7 = 105. That is the mystery number.
Proof: 105 is only divisible by 8 numbers. Start off with 3,5,7 Next three numbers to create are 3*5, 3*7, and 5*7 which are 15,21,35 Last two numbers left are 1 and 105 itself. Adds up to eight numbers! -Alan
Grade 8 POTW: This question wasn't difficult because all you have to do is find the prime factorization of the number, 2000. This will give me 2*2*2*2*5*5*5. I have limits to numbers though! First of all, the number cannot be a decimal since the definition of an integer is: a whole number; a number that is not a fraction.(Thank you, Google.) Second of all, because it says POSITIVE Integer, this means that I have to be above 0. So, using this info, I can create the smallest number! All of these values must be below 10, because if it was 10 or above then it wouldn't be a digit. 5 is the largest digit, because if I multiply it by 2 then it would be 10. So, all 5's stay the same. However, I can multiply 3 2's together, to get 8! Multiplying by another one would give 16, so I can't do that, but it removes two extra digits! So, now I would have 2,8,5,5,5. Since 8 is the largest digit, I can move it to the ones place. And, since 2 is the smallest, I can move it to the leftmost place. This gives me the answer of 25558.
P.S. Another way to remove digits from the 2's were 4 and 4, but I need the smallest ten thousand digit possible, so it would be 2 and 8. The other way would be 44555 as the smallest, but that is not the smallest answer, since 25558 is smaller than that. -Alan
To find the smallest number who's digits multiply up to 2000, the greatest divisor that could divide the dividend without any decimals.
2000/8 =250 250/5 =50 50/5 =2 2/2 1
Once the divisors are found, they are to be arranged in sequence from least to greatest in order to create the smallest number. Because when 1 is multiplied, it does not affect the product so it could be removed and decrease the number by one digit.
To find which digits multiple together to create 2000, all I have to do is find the prime factorization of it. I got 2*2*2*2*5*5*5. I wouldn't want to make the digit 2222555 because I can make it smaller. 2*2*2 is 8. So instead of 2,222,555, I can use 8*2*5*5*5. If I arrange this from least to greatest, I get 25,558.
The smallest possible number that has the digit product of 2000 is 25,558
In order to find the smallest possible number that has the digit product of 2000, we simply have to find the prime factorization of the number, which in this case would be 2*2*2*2*5*5*5. Since this number was already pretty large, I made the number smaller by multiplying the first three 2's together. Put in order least to greatest, that number would be 25,558. Therefore this is my answer!
First step is to gather important information: - Positive number has 8 positive divisors - two of them are 21 and 35 What do I need to find?: the positive number
First of all, I know that the number must be a common multiple of both 21 and 35. I decide to move my way up from the LCM (Lowest common multiple) up to the GCM (greatest common multiple) if it was necessary. I found out that the LCM was 105. From there, I found the divisors of 105, not including 21 and 35. 1, 3, 5, 7, 15, 105 Since that is 6 numbers and including 21 and 35, it makes 8 numbers, I know that the number that I am looking for is 105.
Grade 8 POTW: The smallest number I found was 25558. I used prime factorization like most people, 2^4*5*3=2000, we can multiply 3 twos together to get 8 so we have 2, 5, 5, 5, 8 and there is the answer.
I don't know how to figure a part out, but I think I know the steps; 1. Find what numbers can be diveded by 21 2. Find what numbers can be divided by 35 3. Find a number that can be divided by both numbers 21 and 35 (we are going to subsitute tha number with the variable "n") 4. Find 6 more numbers that have n for the quotient
Following the first example the question gave us with a positive integer of 24, we can use prime factorization to figure out the answer.
2000 / \ 500 4 / \ / \ 250 2 2 2 / \ 25 10 / \ / \ 5 5 5 2 Now we have 2*2*2*2*5*5*5. I'm still a bit confused as to why we're supposed to multiply the first three 2s but not the fourth one. Can someone explain it to me? But anyways, once you multiply the three 2s to get 8, you can arrange the numbers to get 25,558. Therefore, the answer is 25,558. (I tried)
2222555 If you multiply the first 4 2s instead, you'll get 16. 16555 has a digit product of 750, not 2000 (the 16 isn't counted as 16, it's counted as 1 and 6)
The question involves LCM which is lowest common multiple.
We know that the number has 35 and 21 as one of the eight possible digits
21: 3*7 35: 5*7 LCM: 3*5*7 = 105 Eight digits can go into 105 which are 3, 5, 7, 21, 35, 15 and 1 (because one can go into anything) To find the eight digits from the LCM, I just found as many combinations in the sequence (3*5*7)
First I found the lowest common multiple of 35 and 21, it is 105. Knowing this, I would also know that 1 and 105 are divisors. Since I know that 7 is a multiple of 21 which means it is also a divisor. Also, since 3 is a multiple of 21, that means 3 is a multiple of 105. I noticed that since 105 ends with 5, that means 5 is a multiple of 105.
Okie dokies so Basically the question is asking us to find a number that has 8 possible dividers, 2 of which are 21 and 35. Cool cool.
The quickest way to do this is find the lowest common multiple of 35 and 21. 21 and 35 are both divisible by 7. 21 = 3 x 7 35 = 5 x 7
So the least common multiple is 3 x 5 x 7 = 105.
If this answer is correct, that means there are 8 numbers that can divide 105 cleanly.
Obviously, 21 and 35 are two of them. 105 can be divided by 1 and itself, 105 105 can also be divided by 3, 5, 7, and 15. This is a total of 8 divisors.
GRADE 7 POTW: Since two of the divisors are 21 and 35, I would have to find the common divisors of the two numbers. The numbers that both 21 and 35 are divisible by are: 1 and 7. (1 x 21 = 21, 1 x 35 = 35; 3 x 7 = 21, 5 x 7 = 35) The least common multiple of these numbers (3 x 5 x 7) is 105. The divisors of 105 are: 1, 3, 5, 7, 15, 21, 35, and 105. This means that 105 has 8 different divisors, including 21 and 35, making 105 the answer.
The question is asking us to find a number that has 8 possible dividers, and two of which are 21 and 35. To do this, you first have to find the lowest common multiple of 35 and 21.
21 and 35 are both divisible by 7 and 1. 21 = 3 x 7 35 = 5 x 7
As a result, the least common multiple is: 3 x 5 x 7 = 105
If 105 is the correct answer, than, according to the requirements given by the question, it should be divisible by 8 different numbers, and 105 must be divisible by all 8 cleanly.
Obviously, 21 and 35 are two of the 8 different numbers which can divide 105 cleanly. 105 can be also divided by 1, and itself, 105. 105 can also be divided by 3 (35) 5, (21) 7, (15) and 15 (7).
So, the positive integer that fits all the requirements given above would be 105.
Grade 7 POTW:
ReplyDeleteFirst I need to find common multiples with 21 and 35 if I want to find a number that has divisors of 21 and 35. The LCM is 105 (I prime factorize the two numbers and multiply the prime factors together to get 105) Now, I need to factor 105 to see if it has 8 divisors.
105= 1,3,5,7,15,21,35,105
Oh would you look at that, there are 8 positive factors!!!! Oh wow!!!
The answer is 105.
POTW 7:
ReplyDeleteSorry, I forgot to do this POTW so I will do it now (better then never:D)
Info:
- Neither team scored three goals in a row
- There were six goals scored in the match.
- Find all possible scores
- Find how many the Beaver Rangers won
Beaver Rangers: 1 Fox Raiders: 2
Possible Combinations:
1, 1, 2, 1, 1, 2 B
2, 1, 1, 2, 1, 2 Tie
1, 1, 2, 1, 2, 2 Tie
2, 1, 1, 2, 2, 1 Tie
2, 2, 1, 2, 1, 1 Tie
1, 1, 2, 2, 1, 1 B
1, 2, 1, 1, 2, 2 Tie
1, 1, 2, 1, 2, 1 B
1, 1, 2, 2, 1, 2 Tie
1, 2, 1, 1, 2, 1 B
1, 2, 1, 2, 1, 2 Tie
1, 2, 1, 2, 2, 1 Tie
2, 2, 1, 2, 2, 1 F
2, 2, 1, 1, 2, 1 Tie
2, 2, 1, 1, 2, 2 F
1, 2, 2, 1, 1, 2 Tie
1, 2, 2, 1, 2, 1 Tie
2, 1, 2, 1, 1, 2 Tie
2, 1, 2, 1, 2, 1 Tie
1, 2, 2, 1, 2, 2 F
2, 1, 2, 1, 2, 2 F
2, 1, 2, 2, 1, 1 Tie
2, 1, 2, 2, 1, 2 F
2, 1, 1, 2, 1, 1 B
2, 2, 1, 2, 1, 2 F
1, 2, 1, 2, 1, 1 B
There are 26 combinations in which 6 of them are won by the Beaver Rangers.
The mystery number is 105. At first, I multiplied 21 and 35 together to see how many factors would arise from its product. The answer to that was 12. These factors: 1, 3, 5, 7, 15, 21, 35, 49, 105, 147, 245, 735, are all divisors of 735 (21 * 35). Since that didn't work, I decided to find the LCM (lowest common multiple) of the two numbers. Jackpot!!! 105 (the LCM) had exactly 8 positive factors: 1, 3, 5, 7, 15, 21, 35, 105. Seeing as it matches the criteria, I can now be sure that the "integer" is 105.
ReplyDeleteGrade 7 POTW
ReplyDeleteFirst, I'll list the facts
- Integer has exactly 8 divisors
- 2 divisors include 21 and 35
Okay. Knowing that 21 and 35 are a divisor means that a common multiple between the two numbers would result in the number. There are only 8 divisors though so the number won't be too high since the higher you go numerically would most likely have more factors. Also, I know that I only need to find 4 factors because I already somewhat know 4. There is 1, 21, 35, and the number itself.
I'll start with the lowest common multiple.
21, 42, 63, 84, 105, 126, 147...
35, 70, 105, 140...
So far there is 1 common multiple. 105
Factors of 105:
1, 3, 5, 7, 15, 21, 35, 105
105 is the number.
Grade 8 POTW
ReplyDeleteFirst, I'll list the facts.
- Digit product of 2000
- Lowest integer possible
This is going to need some more facts too.
1. Not a 1 or 2 digit because for example, 99 (the highest) has a digit product of 99.
2. No 0s. 0s result in 0 as the product.
3. I need to find all factors of 2000 because no matter what, a number must be a factor of 2000 to have a digit product of 2000.
4. No 2 digit factors of 2000. They will be counted as individual factors
1, 2, 3, 5, 8 are the only digits that are usable (I'm pretty sure), however, others can be used because of multiples and different ways to multiply, and so on, so I'll use them all.
Let's use those to create the integer, but first, I need a way to multiply them to get 2000.
I'm going to see powers of each digit to make it easier.
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048
3, 9, 27, 81, 243, 729, 2187
4, 16, 64, 256, 1024, 4096
5, 125, 625, 3125
6, 36, 216, 1296, 7776
7, 49, 343, 2401
8, 64, 512, 4096
9, 81, 729, 6561
Now, I just need to make 2000. I didn't know why I didn't think of this earlier, but I need to do prime factorization.
2, 2, 2, 2, 5, 5, 5
Now, I will do something that the question didn't say not to do. It's a positive integer. Here is the answer I believe.
1.111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111...
You get the point it continues on infinitely and then the last digits are this:
1111111111111111111111111111111111111111111111111111111111111111111111...2222555.
The digit product is still 2000!
The final lowest integer is 1.111111...2222555.
Umm..... Seayrohn? A decimal is not an Integer.
DeleteYeah. My mistake. Let me re-answer this.
DeleteSame thing as before except the lowest number integer is 2222555. Thanks Alan!
Another mistake. I didn't really "re-do" my answer properly. The old one would've been right if decimals were allowed, but I can't just get rid of the other digits. 2222555 can easily become 25558.
DeleteFinal re-answer is 25558 (unless I mess up again).
test
ReplyDelete21
ReplyDelete35
LCM: 105
105/21= 5
105/35= 3
1,3,5,7,15,21,35,105
There are 8 divisors in 105. The two divisors 21 and 35 are in the factors of 105. I got this by finding the LCM (Lowest Common Multiple) of 21 and 35. The LCM is 105. I then found the factors of 105 which are 1, 3, 5, 7, 15, 21, 35, and 105. There are 8 factors of 105, making it the answer.
I did a similar proccess of Edwyn finding a the smallest common multiple between teh to intergers displayed. I got 105. This even number has 8 divisors and we have listed 2. The other divisors are 5, 1, 15, 35, 105 and 3
ReplyDelete*process
ReplyDeleteFirst I will find the LCM of 21 and 35.
ReplyDelete21: 21, 42, 63, 84, 105
35: 35. 70, 105
The LCM is 105.
The factors/divisors of 105 are 1, 3, 5, 7, 15, 21, 35 and 105.
The integer is 105.
Grade 7 POTW:
ReplyDeleteFirst of all, I would find the least common multiple, but I would rather do it an easier way than guess and check. I prime factorize both numbers, 21 and 35 since the mystery number is divisible by both of them. This is simplified to 3*7 and 7*5. I can get rid of one of the 7's because both numbers have it in common so I only need one. This means that I am left with 3*5*7 = 105. That is the mystery number.
Proof: 105 is only divisible by 8 numbers.
Start off with 3,5,7
Next three numbers to create are 3*5, 3*7, and 5*7 which are 15,21,35
Last two numbers left are 1 and 105 itself.
Adds up to eight numbers!
-Alan
Grade 8 POTW:
ReplyDeleteThis question wasn't difficult because all you have to do is find the prime factorization of the number, 2000. This will give me 2*2*2*2*5*5*5. I have limits to numbers though! First of all, the number cannot be a decimal since the definition of an integer is: a whole number; a number that is not a fraction.(Thank you, Google.) Second of all, because it says POSITIVE Integer, this means that I have to be above 0. So, using this info, I can create the smallest number!
All of these values must be below 10, because if it was 10 or above then it wouldn't be a digit. 5 is the largest digit, because if I multiply it by 2 then it would be 10. So, all 5's stay the same. However, I can multiply 3 2's together, to get 8! Multiplying by another one would give 16, so I can't do that, but it removes two extra digits! So, now I would have 2,8,5,5,5. Since 8 is the largest digit, I can move it to the ones place. And, since 2 is the smallest, I can move it to the leftmost place. This gives me the answer of 25558.
P.S. Another way to remove digits from the 2's were 4 and 4, but I need the smallest ten thousand digit possible, so it would be 2 and 8. The other way would be 44555 as the smallest, but that is not the smallest answer, since 25558 is smaller than that.
-Alan
2222555
ReplyDeletefirst find LCM:
ReplyDelete105
35 x 3 = 105
21 x 5 = 105
then I find all divisors of 105:
1, 3, 5, 7, 15, 21, 35, 105
there are 8 divisors of 105(just my luck), therefore "n" is 105
To find the smallest number who's digits multiply up to 2000, the greatest divisor that could divide the dividend without any decimals.
ReplyDelete2000/8
=250
250/5
=50
50/5
=2
2/2
1
Once the divisors are found, they are to be arranged in sequence from least to greatest in order to create the smallest number.
Because when 1 is multiplied, it does not affect the product so it could be removed and decrease the number by one digit.
Final number: 25,558
Grade 8 POTW:
ReplyDeleteTo find which digits multiple together to create 2000, all I have to do is find the prime factorization of it. I got 2*2*2*2*5*5*5. I wouldn't want to make the digit 2222555 because I can make it smaller. 2*2*2 is 8. So instead of 2,222,555, I can use 8*2*5*5*5. If I arrange this from least to greatest, I get 25,558.
The smallest possible number that has the digit product of 2000 is 25,558
Grade 8 POTW
ReplyDeleteIn order to solve this problem, I found the largest integers under 10 that would divide into 2,000 without the result being a decimal.
2000/8=250
250/5=50
50/5=10
10/5=2
2/2=1
Then, I arranged the digits in order from least to greatest to get 25,558.
Grade 8 POTW:
ReplyDeleteIn order to find the smallest possible number that has the digit product of 2000, we simply have to find the prime factorization of the number, which in this case would be 2*2*2*2*5*5*5. Since this number was already pretty large, I made the number smaller by multiplying the first three 2's together. Put in order least to greatest, that number would be 25,558. Therefore this is my answer!
POTW #8 Grade 7
ReplyDeleteFirst step is to gather important information:
- Positive number has 8 positive divisors
- two of them are 21 and 35
What do I need to find?: the positive number
First of all, I know that the number must be a common multiple of both 21 and 35. I decide to move my way up from the LCM (Lowest common multiple) up to the GCM (greatest common multiple) if it was necessary. I found out that the LCM was 105. From there, I found the divisors of 105, not including 21 and 35.
1, 3, 5, 7, 15, 105
Since that is 6 numbers and including 21 and 35, it makes 8 numbers, I know that the number that I am looking for is 105.
Maxwell's Grade 8 POTW:
ReplyDeleteThe final number would be 25558. I did it in my math notebook
25,558
ReplyDeletemath on paper cause i ran out of notebook space. whoops
The smallest digit I found was 25558.
ReplyDeleteGrade 8 POTW:
ReplyDeleteThe smallest number I found was 25558. I used prime factorization like most people, 2^4*5*3=2000, we can multiply 3 twos together to get 8 so we have 2, 5, 5, 5, 8 and there is the answer.
Bhavees POTS #8
ReplyDeleteI don't know how to figure a part out, but I think I know the steps;
1. Find what numbers can be diveded by 21
2. Find what numbers can be divided by 35
3. Find a number that can be divided by both numbers 21 and 35 (we are going to subsitute tha number with the variable "n")
4. Find 6 more numbers that have n for the quotient
Grade 8 POTW #7
ReplyDeleteFollowing the first example the question gave us with a positive integer of 24, we can use prime factorization to figure out the answer.
2000
/ \
500 4
/ \ / \
250 2 2 2
/ \
25 10
/ \ / \
5 5 5 2
Now we have 2*2*2*2*5*5*5.
I'm still a bit confused as to why we're supposed to multiply the first three 2s but not the fourth one. Can someone explain it to me?
But anyways, once you multiply the three 2s to get 8, you can arrange the numbers to get 25,558.
Therefore, the answer is 25,558. (I tried)
2222555
DeleteIf you multiply the first 4 2s instead, you'll get 16.
16555 has a digit product of 750, not 2000 (the 16 isn't counted as 16, it's counted as 1 and 6)
Oh, okay. Thank you!
DeletePOTW #8:
ReplyDeleteThe question involves LCM which is lowest common multiple.
We know that the number has 35 and 21 as one of the eight possible digits
21: 3*7
35: 5*7
LCM: 3*5*7
= 105
Eight digits can go into 105 which are 3, 5, 7, 21, 35, 15 and 1 (because one can go into anything)
To find the eight digits from the LCM, I just found as many combinations in the sequence (3*5*7)
Therefore, the positive integer is 105
POTW #8
ReplyDeleteFirst I found the lowest common multiple of 35 and 21, it is 105. Knowing this, I would also know that 1 and 105 are divisors. Since I know that 7 is a multiple of 21 which means it is also a divisor. Also, since 3 is a multiple of 21, that means 3 is a multiple of 105. I noticed that since 105 ends with 5, that means 5 is a multiple of 105.
Multiples of 35: 35, 70, 105, 140, 175, 210, 245, 280, 315, 350
Multiples of 21: 21, 42, 63, 84, 105, 126, 147, 168, 189, 210, 231, 252, 273, 294, 315
Grade 8 POTW:
ReplyDeleteI got 25,558. Work in notebook.
21 3 x 7
ReplyDelete35 5 x 7
3
7
5
1
15
A
1 x 3 x 5 x 7 = 105
A = 105
Grade 7 POTW
ReplyDeleteOkie dokies so
Basically the question is asking us to find a number that has 8 possible dividers, 2 of which are 21 and 35. Cool cool.
The quickest way to do this is find the lowest common multiple of 35 and 21.
21 and 35 are both divisible by 7.
21 = 3 x 7
35 = 5 x 7
So the least common multiple is
3 x 5 x 7 = 105.
If this answer is correct, that means there are 8 numbers that can divide 105 cleanly.
Obviously, 21 and 35 are two of them.
105 can be divided by 1 and itself, 105
105 can also be divided by 3, 5, 7, and 15.
This is a total of 8 divisors.
So the positive integer above would be 105.
GRADE 7 POTW:
ReplyDeleteSince two of the divisors are 21 and 35, I would have to find the common divisors of the two numbers.
The numbers that both 21 and 35 are divisible by are: 1 and 7.
(1 x 21 = 21, 1 x 35 = 35; 3 x 7 = 21, 5 x 7 = 35)
The least common multiple of these numbers (3 x 5 x 7) is 105.
The divisors of 105 are: 1, 3, 5, 7, 15, 21, 35, and 105. This means that 105 has 8 different divisors, including 21 and 35, making 105 the answer.
Grade 7 POTW:
ReplyDeleteThe question is asking us to find a number that has 8 possible dividers, and two of which are 21 and 35. To do this, you first have to find the lowest common multiple of 35 and 21.
21 and 35 are both divisible by 7 and 1.
21 = 3 x 7
35 = 5 x 7
As a result, the least common multiple is:
3 x 5 x 7 = 105
If 105 is the correct answer, than, according to the requirements given by the question, it should be divisible by 8 different numbers, and 105 must be divisible by all 8 cleanly.
Obviously, 21 and 35 are two of the 8 different numbers which can divide 105 cleanly.
105 can be also divided by 1, and itself, 105.
105 can also be divided by 3 (35) 5, (21) 7, (15) and 15 (7).
So, the positive integer that fits all the requirements given above would be 105.