Below is this week's POTW. And............HAPPY HALLOWEEN!
Thursday, October 30, 2014
Math POTW #7 - The Search is On
Wow, great work last week Leon D. You were the only student that seemed to obtain the correct answer of 2m. If you got a different answer, please try the problem again and review your work. Do you see why you obtained a different answer? How could you make sure you didn't make a similar error in a future question.
Below is this week's POTW. And............HAPPY HALLOWEEN!
Below is this week's POTW. And............HAPPY HALLOWEEN!
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The answer is 53124. I know this because 3+5=8 and that is divisible by 4. Next I found that 1+2=3 and 3 is obviously divisible by 3. Finally "e' will have to be 4 because it's the last number left.
ReplyDeleteisn't ab a two digit number divisible by 4? not their sum. The same thing for cd
DeleteI think the question is asking you for TWO digit numbers that are divisible by 4 and 3 and they aren't created by adding together. They question says the TWO digit number cd/ab is divisible by 3/4 so it's for example 1 and 5 make 15. 15 is divisible by 3. and 2 and 4 are 24 and is divisible by 4.
DeleteI agree with Leo and Neha, Leon don't you need to find more than one number that satisfies both conditions given?
DeleteTest.
Delete2 clues
ReplyDelete1) The ab is a even number
2) cd is a odd number.
ABCD
13425
1+3=4 4/4=1
4+2=6 6/3=2
and 5 is left
31245
(just the previous number in a different order)
35421
3+5=8 8/4=2
4+2=6 6/3=2
and 1 is left
53241
(same as the previous number with a different order)
53124
5+3=8 8/4=2
1+2=3 3/3=1
and 4 is left
35214
(same as the previous number with a different order)
Since ab is divisible by 4 and only numbers, 1,2,3,4, and 5 can be used, the only possible combinations are 12, 24 and 32. Since cd is divisible by 3, the only combinations are 12,15,21 and 54.
ReplyDeleteI tried each combinations for the first 4 numbers as it is abcd without repeating numbers:
I got the last digit for each as it is the only one left
12543
24153
32154
32541
Therefore, there are 4 answers to this question 12543, 24153, 32154 and 32541.
The answer is 24153
ReplyDeleteab is 24. 24 is divisible by 4 (24/4 = 6)
cd is 15. 15 is divisible by 3 (15/3 = 5)
The left over number (assuming it is "e") is 3.
the possible digits for ab are
ReplyDelete12, 24, 32, 52
the possible digits for cd are
12, 15, 21, 24, 42, 45
so the possible combinations of ab and cd are
1245, 3215, 2415, 3245
therefore the possible numbers that could be formed from the digits 5,4,3,2,1 are...
12453, 24153, 32154, and 32451
12, 24, and 32 are divisible by 4, and 45 and 15 are divisible by 3 (12/3 is 4, 24/6 is 4, 32/8 is 4, 45/15 is 3, and 15/5 is 3).
ReplyDelete12453, 24153, 32154, and 32451 are all the five digit numbers that satisfy both conditions since all the digits in every five-digit are not the same.
53124: The old answer where I thought that it was ab=a+b
ReplyDelete24153: The new answer where I put a number beside another to create a double digit number
First I picked 2 random numbers that put together could be divided by 4. I picked 1 and 2 making 12. 12 divided by 4 is 3. The leftover numbers are 3, 4, 5. Again I picked two random numbers that could be divisible by 3. I picked 4 and 5 making 45. 45 divided by 3 is 15. ab (12) and cd (45) put together would be 1245. Since there is only 1 number left (3). THis number had to be e. So I added it to the end of abcd. This made the answer 12453.
ReplyDeleteFirst I picked 2 random numbers that put together could be divided by 4. I picked 1 and 2 making 12. 12 divided by 4 is 3. The leftover numbers are 3, 4, 5. Again I picked two random numbers that could be divisible by 3. I picked 4 and 5 making 45. 45 divided by 3 is 15. ab (12) and cd (45) put together would be 1245. Since there is only 1 number left (3). THis number had to be e. So I added it to the end of abcd. This made the answer 12453.
ReplyDeleteCalista wrote:
ReplyDeletePossible numbers for ab:
12, 24, 32
Possible numbers for cd:
12, 15, 21, 24, 42, 45, 51, 54
Numbers formed:
12453, 12543, 24153, 24513, 32154, 32451, 32514, 32541
Therefore the numbers formed that satisfy both conditions are: 12453, 12543, 24153, 24513, 32154, 32451, 32514, and 32541.
To answer this question, i laid out all the numbers that are multiples of 3&4 that had digits 1-5
ReplyDelete3 4
12,15,21,24,33,42,45,51,54 12,24,32,44,52
I could get rid of 44 and 33 as u could only use a number once. Then i found a combination of the new numbers that had completely different digits. This got me 15324, as 15 was a multiple of 3 and 32 was a multiple of 4, and 4 was added on as u needed to use all digits.45321, and 45123 would work as well since 45 is a multiple of 3 and 32/12 are multiples of 4. There are many other answers as well since there are many numbers out of the ones i listed above which can pair with another number from the other group.
numbers divisible by 3(cd)= 12 15 24 45 54 42 51
ReplyDeletenumbers divisible by 4(ab)= 12 24 52 32
12345, 12354, 12342, 24315 24351, 32415, 32145, 32154, 32451
i went along my key board from 1-5 like this 12131415,232425,3435,45
and them back wards 54535251,434241,3231,21
and saw which were divisible by 3 and then i did four
after i put the numbers together that's what i got
Suren Wrote:
ReplyDeleteNumbers:1 2 3 4 5
ab divisible by 4
cd divisible by 3
I picked random numbers that worked for the criteria. When I thought of the first criteria, I came up with 12. 12 was divisible by 4. Then, I had numbers 3,4, and 5 left. Since I had to make a number divisible by 3, it didnt work. Then I went to the next set of numbers for the criteria. 24. 24 was divisible by 4. After this, I would have numbers 1,3, and 5 left. 15 is divisible by 3. After this, the only number left was 3, so the 5 digit number I came up with is 24153
Aparna wrote:
ReplyDeleteFirst I found the possible numbers for ab:
12, 24, 32
the possible numbers for cd are:
12,15,21,24,42,45,51,54
I know that the numbers for cd cannot have a two because the possibilities for ab all have 2
so it can only be:
15, 45, 51, 54
the numbers formed are: 12345, 12354, 24315, 24351, 32415, 32145, 32451, 32154
therefore there are 8 possible combos that meet the conditions and they are 12345, 12354, 24315, 24351, 32415, 32145, 32451, and, 32154.
Hope wrote:
ReplyDeleteYou can make four numbers this way, 12435, 24315, 32154, and 32514. I figured this out by using skip counting to first figure out the first digits divisible by 4 and then used mental math and common sense to tell if they were able to be put in the context of the question. (only using numbers from 1-5) I also used the same strategies to tell if I could make the next number divisible by 3 with the numbers I had left. Eventually I came to the the number 52 when finding a number divisible by 4. Though when using this number I couldn’t find a number divisible by 3, I did know that this was the last possible set (or sets that may start with 52) that I could use since the next number divisible by 4 was 56, and then 60 which wouldn’t fit the criteria of the question. (And onward)
abcde=24153
ReplyDeleteI used guess and check to find out the answer
I tried 24 for the first two numbers because I knew that this was divisible by 4 and I also didnt want to use 1 and 2 in the first two digits because it would be impossible to do the others.
I did 15 and not 13 for the third and fourth digits because I knew 13 is a prime number and 15 is divisble by 3.
All possible 2 digit combinations:
ReplyDelete12, 13, 14, 15, 21, 23, 24, 25, 31, 32, 34, 35, 41, 42, 43, 45, 51, 52, 53, 54
Numbers divisible by 4:
12, 24, 32, 52
Numbers divisible by 3:
12, 15, 21, 24, 42, 45, 51, 54
Combinations from these:
1245, 1254, 2415, 2451, 3215, 3245, 3251, 3254, 52 N/A
Then, I just add the missing number to each one
All possible combinations are:
12453, 12543, 24153, 24513, 32154, 32451, 32514, 32541
Many students got VERY close (Aparna, Dan...) however no one got the perfect response of 8 numbers:
ReplyDelete12453, 12543, 24153, 32154, 32514, 32451, and 32541. If you obtained some different version of these 8 numbers, can you see why? Are you able to edit your process?
OMG! I am so sorry to Dan! You actually got the correct answer. That is awesome-o!
ReplyDeleteabcde=24153
ReplyDeleteI made a chart and then used guess and check