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Info: - 1,2,3,4,5 are used to make number abcdef. -abc is odd -bcd/5 -cde/3 Since abc is odd, c must be 1,3 or 5. d must be 5 because it is the only number divisible by 5, meaning that c cannot be 5 because there is no other possibility for d. Possibilities for c and e must be be either 1 or 3. Because cde is divisible by 3, and 1(c)+5(d)=6, e must be 3. And because 3(c)+5(d)=8, e must also be 1. c and e can both use th e. Possibilities for a and b can be anything else other than the numbers already used. Possibilities for a: 2,4 Possibilities for b: 2,4 Possibilities for c: 1,3 Possibilities for d: 5 Possibilities for e: 1,3 All possibilites include: - 24153 - 24351 - 42351 - 42153
Grade 7/8 POTW I will start my acknowledging the simple and straightforward info based on this question. Since you can only use 1,2,3,4 and 5 in the number, C must be either 1,3, or 5 since they are all odd. Looking at the second aspect of the question when it says BCD is divisible by 5, D has to be 5 and that is the only possibility for it. This means that C can only be 1 or 3. Also, if CDE is divisible by 3, then e should be 1 or 3 because if you add up all of the numbers in CDE, it should equal to 9. 1+5+3=9/3=3. 3 is a valid possibility for E. 3+5+1=9/3=3. 1 is also a valid possibility for E. I have now found all possibilities for D, C and E meaning that A and B can be any of the other numbers. A can be 2 or 4 B can be 2 or 4 C can be 1 or 3 D can only be 5 E can be 1 or 3. Now I will find all possibilities of a 5 digit number corresponding with these rules. - 42153 - 42351 - 24153 - 24351 All I had to do was switch around 2 and 4 after I switched 1 and 3 with C and E. There are only 4 possibilities for the 5 digit number if I did my math correctly.
(Ethan) The rules are: 1. Digits ABC (from ABCDE) must be odd, so C must be either 1, 3, or 5. 2. Digits BCD (from ABCDE) must end in 5 or 0, so the only number D can suit is 5. Therefore, D = 5 3. Digits CDE from ABCDE) must be divisible by 3. Now, we have a solution. The digits can be: (after hard work done) 42351 (423 = odd, 235 is divisible by 5, and 351 is divisible by 3) Now you can switch C and E because either digit can be either 1 or 3, and the result would still work. 42153 (423 = odd, 215 = still divisible by 5, and 153 = divisible by 3) Digits A and B can now only be 2 or 4 because B, D, and E cannot be 2 or 4. However, like C and E, they can be switchabe because it works either way. 24153 (241 = still odd, 415 = still divisible by 5, 153 = divisible by 3) Finally, 24351 (243 = odd, 435 = divisible by 5, 351 = divisible by 3). So, the possibilities include: 42351 42153 24153 24351
Maruti Since BCD is divisible by 5 and no letter is 0, it means that D = 5. C is odd so it is either 1 or 3. E can also be either 1 or 3 and A and B can be 2 or 4.
Since CDE is divisible by 3, it has to equal a multiple of 3. 42,351 42,153 24,153 24,351 just switch the digits 3 and 1, and 2 and 4 to get all the possibilities. It could also be 12,354 and 21,354 because the last 3 digits equal 12 and the first digits don't have any rules.
Grade 7/8 POTW In this question, we have to use the digits 1-5 once to create a five digit number, abcde that follows certain requirements: - abc is odd - bcd is divisible by 5 - cde is divisible by 3 To start things off, each variable from a-e can be digits 1-5. If abc is odd, it means that c must be either 1, 3, or 5 because all odd numbers end with 1, 3, 5, 7, or 9. If bcd is divisible by 5, it means that it must be 5 because all numbers divisible by 5 must end in either 5 or 0 and 0 isn't an option. All numbers divisible by 3 can be any digit. With the new information, we know that a can be 1-4, b can be 1-4, c can be 1 or 3 (5 is taken already), d is 5, and e can be 1-4. Now, we can focus on cde. All numbers divisible by 3 must have their digits adding up to a number divisible by 3. We know cde is c5e. We also know c5e can be 15e or 35e. In 15e, the only other value of e can be 3. In 35e, e can be 1 or 4. That means abcde can be the following: ab153 ab351 ab354 Due to the fact that a and b don't have any requirements except for not being the same as the other variables, the values of both can be 1-4 or any of the available digits. The possibilities include: 24153 24351 21354 42153 42351 12354 Therefore, there are 6 possible abcde values.
Aliyah: Since the bcd must be divisible by 5 and there is no zero in the allowed numbers d must be 5 for ALL possibilities. And since abc must be odd and 5 is already taken c must be either 1 or 3. That means that with multiple possibilities it can be either 1 or 3. Lastly since cde must be divisible by 3 cde's digits must add up to a multiple of 3. THEREFORE................. Possibilities include: 24153 24351 12354 21354 42153 42351
I started off trying to find cde. To find cde, I used trial and error to find 3 number that would be a multiple of three. Each digit that I used to add is c + d + e = multiple of 3. All of the possible numbers for cde are listed below. 5 + 4 + 3 = 12 5 + 3 + 4 = 12 4 + 5 + 3 = 12 4 + 3 + 5 = 12 3 + 4 + 5 = 12 3 + 5 + 4 = 12 5 + 3 + 1 = 9 5 + 1 + 3 = 9 3 + 5 + 1 = 9 3 + 1 + 5 = 9 1 + 3 + 5 = 9 1 + 5 + 3 = 9 3 + 2 + 1 = 6 3 + 1 + 2 = 6 2 + 1 + 3 = 6 2 + 3 + 1 = 6 1 + 2 + 3 = 6 1 + 3 + 2 = 6 4 + 3 + 2 = 9 4 + 2 + 3 = 9 3 + 4 + 2 = 9 3 + 2 + 4 = 9 2 + 3 + 4 = 9 2 + 4 + 3 = 9 Now, we can take out all the numbers in which d is not 5 or 0, as it must be a multiple of 5. The remaining numbers are: 453 354 351 153 We can take away 453, as c isn’t an odd number This leaves a and b to be filled by either 2, 4, or 1 Therefore, the possible numbers are: 12354 21354 24153 24351 41354 42153
Another way to solve this problem would be: For the 5 digit number made of 1,2,3,4,5 to satisfy all the 3 conditions : d is always 5 ( only no that is divisible by 5) c is always 1 or 3 ( for abc to be odd) e can be either 1 or 3 or 4 a and b to be filled by either of 2 or 4 or 1 Therefore, the possible numbers are: 12354 21354 24153 24351 41354 42153
For the number divisible by 5, (d) is always 5. (c) is always odd (1 or 3) because it is at the end of an odd number, and therefore is odd. Because we know the first two digits of cde are 3 and 5 or 1, we know that e has to be 3, 4 or 1. (351 and 153). The second and first number can be swapped interchangeably, and thus can be 1, 2, 3 or 4 (however, one of them HAS to be a two!) Possibilities: 12354 21354 24153 42153 42351
Info:
ReplyDelete- 1,2,3,4,5 are used to make number abcdef.
-abc is odd
-bcd/5
-cde/3
Since abc is odd, c must be 1,3 or 5. d must be 5 because it is the only number divisible by 5, meaning that c cannot be 5 because there is no other possibility for d. Possibilities for c and e must be be either 1 or 3. Because cde is divisible by 3, and 1(c)+5(d)=6, e must be 3. And because 3(c)+5(d)=8, e must also be 1. c and e can both use th e. Possibilities for a and b can be anything else other than the numbers already used.
Possibilities for a: 2,4
Possibilities for b: 2,4
Possibilities for c: 1,3
Possibilities for d: 5
Possibilities for e: 1,3
All possibilites include:
- 24153
- 24351
- 42351
- 42153
My POTW was done on paper: some possibilities include:
ReplyDelete42153
24351
34152
42351
24153
Grade 7/8 POTW
ReplyDeleteI will start my acknowledging the simple and straightforward info based on this question. Since you can only use 1,2,3,4 and 5 in the number, C must be either 1,3, or 5 since they are all odd. Looking at the second aspect of the question when it says BCD is divisible by 5, D has to be 5 and that is the only possibility for it. This means that C can only be 1 or 3. Also, if CDE is divisible by 3, then e should be 1 or 3 because if you add up all of the numbers in CDE, it should equal to 9. 1+5+3=9/3=3. 3 is a valid possibility for E. 3+5+1=9/3=3. 1 is also a valid possibility for E. I have now found all possibilities for D, C and E meaning that A and B can be any of the other numbers.
A can be 2 or 4
B can be 2 or 4
C can be 1 or 3
D can only be 5
E can be 1 or 3.
Now I will find all possibilities of a 5 digit number corresponding with these rules.
- 42153
- 42351
- 24153
- 24351
All I had to do was switch around 2 and 4 after I switched 1 and 3 with C and E. There are only 4 possibilities for the 5 digit number if I did my math correctly.
(Ethan)
ReplyDeleteThe rules are:
1. Digits ABC (from ABCDE) must be odd, so C must be either 1, 3, or 5.
2. Digits BCD (from ABCDE) must end in 5 or 0, so the only number D can suit is 5.
Therefore, D = 5
3. Digits CDE from ABCDE) must be divisible by 3.
Now, we have a solution.
The digits can be:
(after hard work done)
42351 (423 = odd, 235 is divisible by 5, and 351 is divisible by 3)
Now you can switch C and E because either digit can be either 1 or 3, and the result would still work.
42153 (423 = odd, 215 = still divisible by 5, and 153 = divisible by 3)
Digits A and B can now only be 2 or 4 because B, D, and E cannot be 2 or 4.
However, like C and E, they can be switchabe because it works either way.
24153 (241 = still odd, 415 = still divisible by 5, 153 = divisible by 3)
Finally,
24351 (243 = odd, 435 = divisible by 5, 351 = divisible by 3).
So, the possibilities include:
42351
42153
24153
24351
D has to be a 5
ReplyDeleteC has to be a 1 or a 3
c+d+e=9 or a 12
Possible numbers for cde
cde= 153
cde= 351
cde= 354
So the total possible combinations are...
24153
42153
42351
24351
21354
12354
6 numbers in total that follow the instructions
Maruti
ReplyDeleteSince BCD is divisible by 5 and no letter is 0, it means that D = 5.
C is odd so it is either 1 or 3.
E can also be either 1 or 3 and A and B can be 2 or 4.
Since CDE is divisible by 3, it has to equal a multiple of 3.
42,351
42,153
24,153
24,351
just switch the digits 3 and 1, and 2 and 4 to get all the possibilities.
It could also be 12,354 and 21,354 because the last 3 digits equal 12 and the first digits don't have any rules.
Grade 7/8 POTW
ReplyDeleteIn this question, we have to use the digits 1-5 once to create a five digit number, abcde that follows certain requirements:
- abc is odd
- bcd is divisible by 5
- cde is divisible by 3
To start things off, each variable from a-e can be digits 1-5. If abc is odd, it means that c must be either 1, 3, or 5 because all odd numbers end with 1, 3, 5, 7, or 9. If bcd is divisible by 5, it means that it must be 5 because all numbers divisible by 5 must end in either 5 or 0 and 0 isn't an option. All numbers divisible by 3 can be any digit.
With the new information, we know that a can be 1-4, b can be 1-4, c can be 1 or 3 (5 is taken already), d is 5, and e can be 1-4. Now, we can focus on cde. All numbers divisible by 3 must have their digits adding up to a number divisible by 3. We know cde is c5e. We also know c5e can be 15e or 35e. In 15e, the only other value of e can be 3. In 35e, e can be 1 or 4. That means abcde can be the following:
ab153
ab351
ab354
Due to the fact that a and b don't have any requirements except for not being the same as the other variables, the values of both can be 1-4 or any of the available digits. The possibilities include:
24153
24351
21354
42153
42351
12354
Therefore, there are 6 possible abcde values.
Aliyah:
ReplyDeleteSince the bcd must be divisible by 5 and there is no zero in the allowed numbers d must be 5 for ALL possibilities. And since abc must be odd and 5 is already taken c must be either 1 or 3. That means that with multiple possibilities it can be either 1 or 3. Lastly since cde must be divisible by 3 cde's digits must add up to a multiple of 3. THEREFORE.................
Possibilities include:
24153
24351
12354
21354
42153
42351
I started off trying to find cde. To find cde, I used trial and error to find 3 number that would be a multiple of three.
ReplyDeleteEach digit that I used to add is c + d + e = multiple of 3. All of the possible numbers for cde are listed below.
5 + 4 + 3 = 12
5 + 3 + 4 = 12
4 + 5 + 3 = 12
4 + 3 + 5 = 12
3 + 4 + 5 = 12
3 + 5 + 4 = 12
5 + 3 + 1 = 9
5 + 1 + 3 = 9
3 + 5 + 1 = 9
3 + 1 + 5 = 9
1 + 3 + 5 = 9
1 + 5 + 3 = 9
3 + 2 + 1 = 6
3 + 1 + 2 = 6
2 + 1 + 3 = 6
2 + 3 + 1 = 6
1 + 2 + 3 = 6
1 + 3 + 2 = 6
4 + 3 + 2 = 9
4 + 2 + 3 = 9
3 + 4 + 2 = 9
3 + 2 + 4 = 9
2 + 3 + 4 = 9
2 + 4 + 3 = 9
Now, we can take out all the numbers in which d is not 5 or 0, as it must be a multiple of 5. The remaining numbers are:
453
354
351
153
We can take away 453, as c isn’t an odd number
This leaves a and b to be filled by either 2, 4, or 1
Therefore, the possible numbers are:
12354
21354
24153
24351
41354
42153
Another way to solve this problem would be:
ReplyDeleteFor the 5 digit number made of 1,2,3,4,5 to satisfy all the 3 conditions :
d is always 5 ( only no that is divisible by 5)
c is always 1 or 3 ( for abc to be odd)
e can be either 1 or 3 or 4
a and b to be filled by either of 2 or 4 or 1
Therefore, the possible numbers are:
12354
21354
24153
24351
41354
42153
For the number divisible by 5, (d) is always 5. (c) is always odd (1 or 3) because it is at the end of an odd number, and therefore is odd. Because we know the first two digits of cde are 3 and 5 or 1, we know that e has to be 3, 4 or 1. (351 and 153). The second and first number can be swapped interchangeably, and thus can be 1, 2, 3 or 4 (however, one of them HAS to be a two!)
ReplyDeletePossibilities:
12354
21354
24153
42153
42351
Done on Paper
ReplyDelete42153
24351
34152
42351
24153