POTW #4 Solution Gr. 7:
Please refer to Anya's calculations among many others who found the correct measures of central tendency. As for which salary would MOST likely be Aniya's...please refer to RWA reasonings. This means there is a real-world answer. In MOST business or places of employment there is a pyramid of power/responsibility (think of the CEO, to the board of executives, to the managers, then to the workers). In MOST cases regardless of how long you've been employed, those on the higher levels tend to make a higher salary.
POTW #4 Solution Gr. 8:
Please refer to Fiona or Seayrohn's calculations and discussion.
POTW #5 Question: (National Palindrome Day)
There are multiple six-digit palindrome numbers divisible by 15.
ReplyDelete- 513315
- 516615
- 519915
- 522225
- 525525
- 528825
- 531135
- 534435
- 537735
- 540045
- 543345
- 546645
- 549945
- 552255
- 555555
- 558855
- 561165
- 564465
- 567765
- 570075
- 573375
- 576675
- 579975
- 582285
- 585585
- 588885
- 591195
- 594495
- 597795
I know that each of these work and each are divisible by 15 because it must start with 5 and end with 5 because 15 can only go into something that ends in either 5 or 0 and since it's a palindrome it must only be 5 as you can't start it with 0 otherwise it's just a 5 digit number. As well, another trick is to also add 3 to the 3rd and 4th digit in the number and you will get all 30
In total there are 30 total different palindromes divisible by 15.
This comment has been removed by the author.
DeleteSorry, can't remember if I put it or not but there is also 510015 making it 31. Sorry, I forgot to add because there were so many and it was a long long list
ReplyDeleteMaruti Singh
ReplyDeleteSo first we should find the number of palindromes, first we know it should start at a hundred thousand digit of 5 so it is at least divisible by 15. The first number that follows the rules is 510,015. The next is 513,315, the next is 516,615. Then, a pattern occurs, add 3,300 to find the next number. Then I made a list of all the numbers.
510,015
513,315
516,615
519,915 Then it increases by 2310.
522,225
525,525
528,825 Then it increases by 2310.
531,135
534,435
537,735
540,045 Increases by 2310
543,345
546,645
549,945
552,255 Increases by 2310
555,555
558,855
561,165 Increases by 2310
564,465
567,765
570,075 Increases by 2310
573,375
576,675
579,975
582,285 Increases by 2310
585,585
588,885
591,195 Increases by 2310
594,495
597,795
There are 30 6 digit palindromes that are divisible by 15.
Sorry this is the last time, I think I actually did 510015. But I honestly have no clue. Sorry!
ReplyDeletePOTW
ReplyDeleteThere are 900 total 6 digit palindromes as the first digit has to be something other than 0 so 1-9 and the next 2 can be from 0-9. Since the last 3 digits are the same, we have to multiply all of the possibilities for digits. 9 times 10 times 10=900. There are 900 6 digit palindromic numbers. Now, in order to find out the number of palindromic numbers divisible by 15, we have to divide 900 by 15. 900/15=60.
There are 60 6 digit palindromic numbers that are divisible by 15.
Grade 7/8 POTW:
ReplyDeleteKnowing that palindromes are the same forwards and back, I know that once you choose the first 3 numbers, you don't have the freedom to choose the last 3. So, for the first 3 numbers, the first one cannot be a 0, so it can be anything from 1-9. The other 2 can be whatever. 9 x 10=900 total 6 digit palindromic numbers. And to find the number of those 900 that are divisible by 15, you divide 900 by 15. 900/15=60. There are 60 six digit palidromes that are divisible by 15.
(Forgot to add this): We have to divide by 15 because every 15 numbers, there is a number divisible by 15.
DeleteSorry, I realized that my answer is actually wrong. I think it would be better if I listed all the palindromes instead. To help me narrow this down even more, I know that the first three numbers must add up to a multiple of 3, starting from 6 and ending at 21 (5+9+7).
Delete501105 (1)
504405 (2)
507705 (3)
510015 (4)
513315 (5)
516615 (6)
(At this point, I noticed that the third digit must increase my 3 each time until we reach the point where we have to change the first digit. Before this point, I manually added the digits until I found a number that satisfied what we were finding)
519915 (7)
522225 (8)
525525 (9)
528825 (10)
531135 (11)
534435 (12)
537735 (13)
540045 (14)
543345 (15)
546645 (16)
549945 (17)
552255 (18)
555555 (19)
558855 (20)
561165 (21)
564465 (22)
567765 (23)
570075 (24)
573375 (25)
576675 (26)
579975 (27)
582285 (28)
585585 (29)
588885 (30)
591195 (31)
594495 (32)
597795 (33)
Aliyah,
ReplyDeleteIn total there are 900 6 digit Palindromic numbers. 60 of which are divisible by 15.
The first number cannot be zero so it must be from 1-9. Next the other numbers can be anything.(remember once you choose the first 3 numbers you can't choose anymore.) Therefore 9x10=900. To find out how many are divisible by 15, you need to divide 900/15. This is because every 15 numbers one I divisible by 15. There are 60 palindromic numbers divisible by 15.
I knew that for a number to be divisible by 15 it has to be divisible by 3 and 5. Therefore all of the numbers had to start with a 5 since a number can't start with a 0. Also the first three numbers had to add up to- 6, 9, 12, 15, 18, 21. That is because if I add the last three numbers to that I would get- 12, 18, 24, 30, 36, 42. That procedure told me that the number is also divisible by 1 and 5 (15).
ReplyDelete1)501105
2)504405
3)507705
4)510015
5)513315
6)516615
7)519915
8)522225
9)525525
10)528825
11)531135
12)534435
13)537735
14)540045
15)543345
16)546645
17)549945
18)552255
19)555555
20)558855
21)561165
22)564465
23)567765
24)570075
25)573375
26)576675
27)579975
28)582285
29)585585
30)588885
31)591195
32)594495
33)597795
There are 33 different 6-digit palindromes that are divisible by 15.
Grade 7/8 POTW
ReplyDeleteWhat you have to do is find out how many six digit palindromes are divisible by 15. In order to do this, first you have to figure out how many palindromes there actually are. The first 3 digits are just repeated in reverse as the last 3 digits so we just have to identify the first 3. 0 can't begin a number (except for 0), so there are 1-9 or 9 possible digits for the first digit. The next 2 can however be 0 and 1-9. This means there are 9 x 10 x 10 possible 6 digit palindromes. In order to find out how many of these are divisible by 15, there are 2 ways. One way is somewhat correct and incorrect (in other words it is incorrect) as it doesn’t properly display the full answer while the other one does. The first way involves dividing 900 by 15 to get 900/15 = 60 possible 6 digit palindromes. However, we still aren't done. It's hard to explain this, but the number 900 doesn't recognize the fact that the palindromes MUST start and end with a 5. This is because multiples of 15 ALWAYS end in either 5 or 0 and if it ends in 0, it'll have to begin with 0. Therefore, it can only end with 5. We have to include the fact that multiples of 30 are included since 900/30 = 30 which still works even though there should be 0. In other words, 15 x 2 = 30 so we have to divide 900/15/2 in order to get the “real answer” (this entire process is really generally incorrect, I’m just showing how and why). 900/15 = 60, 60/2 = 30 (This is actually still not the right answer). That was the first improper way. The second way is to actually take the total amount of palindromes (900) and divide it by 9. This is because we already established that the first digit is 5 so the rest actually negatively affects the data. 900/9 = 100. From there, we can divide 100 by 3. You may be confused of why divide by 3, but this is important. Since we already divided 900 by 9, we make it so that all the 100 palindromes are divisible by 5 (it all ends in 5 now). 5 x 3 = 15 so by eliminating the requirement of dividing by 5, we just need to divide by 3 to get the total amount. 100/3 = 33.3333333... This is okay because there is actually an extra multiple of 5 in the 100 palindromes (67). I’m not going to explain too much on this (don’t want to go on forever). Anyways, 33.333333… is rounded (and actually) to 33 meaning there are 33 6-digit palindromes divisible by 15. .To prove this, we can also list the possibilities. (NOTE: IT MUST START AND END WITH 5).
510015
540045
570075
501105
531135
561165
591195
522225
552255
582285
513315
543345
573375
504405
534435
564465
594495
525525
555555
585585
516615
546645
576675
507705
537735
567765
597795
528825
558855
588885
519915
549945
579975
Therefore, there are 33 6-digit palindromes that are divisible by 15.
5x3=15
ReplyDeleteIf you want the number to be dividable by 5 it has to end in five or zero. It can't start in zero so there for the last number can't be zero. We have determined that the first and last number has to 5 to be divisible by 15. Now that we have eliminated the 5 the rest of the numbers has to be divisible by 3. So if we were to add all the digits they would have to be divisible by 3, because we have established that the first and last numbers are 5 the middle numbers would have to add up to: 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32 and 35, to be divisible by 3. And we can eliminate the odd numbers because it won't add up. So we are left with 2, 8, 14,20, 26 and 32 being the four middle numbers.
Middle numbers equaling 2
- 510015
- 501105
Middle numbers equaling 8
- 540045
- 504405
- 513315
- 531135
- 522225
Middle numbers equaling 14
- 570075
- 507705
- 516615
- 561165
- 525525
- 552255
- 534435
- 543345
Middle numbers equaling 20
- 591195
- 519915
- 528825
- 582285
- 537735
- 573375
- 546645
- 564465
- 555555
Middle numbers equaling 26
- 549945
- 594495
- 558855
- 585585
- 576675
- 567765
Middle numbers equaling 32
- 579975
- 597795
- 588885
In total there are 33 6-digit palindromes that are divisible by 15
If you want a palindrome number to be divisible by 15, it must have a digit root of 15. The highest 6 digit palindrome number is 999999, which has a digit root of 54. Therefore, the numbers that the palindrome can add up to are as follows:
ReplyDelete15, 30, 45
We can immediately cross out 45 and 15, since they are not possible to add up to with sums of two numbers. This is because the sum of two numbers is always an even number, and therefore, if you are adding a palindrome, the resultwould always equal an even number. (eg. 501105's digit root (even though the number is odd) is 12.) Therefore, the palindrome can only add up to 30.
Now, we need to determine numbers that add up to 30. If the first and last numbers are 0, we need the middle 4 to add up to 30. If the first and last numbers are 1, we need to add up to 28, and so on. That leaves us with these numbers:
30, 28, 26, 24, 22, 20, 18, 16, 14, 12
Looking at how many number sums ch number has, a pattern quickly becomes apparent.
For example, adding up to thirty the number sums are 14 and 16, along with 12 and 18. You can also reverse these to get different number sums (e.g. 078870 and 087780.) Therefore, we need only to look at the number sums of each number to find our answer. (And it will be even, because we are counting by multiples of 2.)
Second part coming
Number sums of 30: {12, 18} {18, 12}
ReplyDelete*{} means a number that can be flipped, [] indicates a number cannot.
Number sums of 28:[14, 14] {12, 16} {10, 18}
Number sums of 26:{12, 14} {10, 16} {8, 18}
Looking at these, we see a pattern. It's a staircase!
4,5,6... and so on! But where does the staircase end? To find this out, we need to look at the last number sum, which is 12. We get:
[6,6] {4,8} {2,10} {0, 12}
giving us 7 number sums. The staircase, then, looks like this:
4,5,6,7,8,9,10,9,8,7
4 number sums of 30
5 number sums of 28
.
.
.
and so on. Adding these all together, we get:
4+5+6+7+8+9+10+9+8+7 = 73 palindromes. (it wasn't an even number because the staircase ended on a 7)
The End!
(in total there are [I think anyway] 73 6 digit palindromes divisible by 15)
To solve this problem with guess and check we first need to know one thing, numbers divisible by 3 and 5 are divisible by 15 and, that palindromes are able to be read backwards the same way as forwards like "race car".
ReplyDeletethe palindromes are:
- 510015
- 501105
- 540045
- 504405
- 513315
- 531135
- 522225
- 570075
- 507705
- 516615
- 561165
- 525525
- 552255
- 534435
- 543345
- 591195
- 519915
- 528825
- 582285
- 537735
- 573375
- 546645
- 564465
- 555555
- 549945
- 594495
- 558855
- 585585
- 576675
- 567765
- 579975
- 597795
- 588885
When we add them all up there are a total of 33, therefore, there are a total of 33 palindrome divisible by 15 and have 6 digits.
First I knew that the number had to end with five because the number had to be divisible by 15 then the first number has to be 15 because it’s a palindrome then you find the multiples of 15 that are palindromes. I noticed from the thousands and hundreds that they kept going up by 3 until they were changing the ten- thousands place, then they increased by 4.
ReplyDelete501105
504405
507705
510015
513315
516615
519915
And so on until you reach a total of 33 numbers
First I knew that the number had to end with five because the number had to be divisible by 15 then the first number has to be 15 because it’s a palindrome then you find the multiples of 15 that are palindromes. I noticed from the thousands and hundreds that they kept going up by 3 until they were changing the ten- thousands place, then they increased by 4.
ReplyDelete501105
504405
507705
510015
513315
516615
519915
And so on
First I knew that the number had to end with five because the number had to be divisible by 15 then the first number has to be 15 because it’s a palindrome then you find the multiples of 15 that are palindromes. I noticed from the thousands and hundreds that they kept going up by 3 until they were changing the ten- thousands place, then they increased by 4.
ReplyDelete501105
504405
507705
510015
513315
516615
519915
And so on until they reach 33 possible palindromes, if decimals count as digits then just multiply the 6 digits by 33 to get 198
First I knew that the number had to end with five because the number had to be divisible by 15 then the first number has to be 15 because it’s a palindrome then you find the multiples of 15 that are palindromes. I noticed from the thousands and hundreds that they kept going up by 3 until they were changing the ten- thousands place.
ReplyDelete501105
504405
507705
510015
513315
516615
519915
522225
And so on until they reach 33 possible palindromes, if decimals count as digits then just multiply the 6 digits by 33 to get 198
I am going to solve this by narrowing down what I need to do to make it easiest for myself. Starting off, I know that if a number is divisible by 15, it must end in a 5 or a 0. I also know for it to be a 6 digit number, it can’t start with 0. This means that every number starts in a 5. For it to be a palindrome, it also ends in a 5. From here I just need to find the letters in for: 5abba5. I also know that 5+a+b must equal something divisible by 3. From here, I just tried number combinations. The palindromes I found were:
ReplyDelete501105
504405
507705
510015
513315
516615
519915
522225
525525
528825
531135
534435
537735
540045
543345
546645
549945
552255
555555
558855
561165
564465
567765
570075
573375
576675
579975
582285
585585
588885
591195
594495
597795
There are 33 possible 6 digit palindrome combinations.
In order to have a 6 digit number divisible by 15, it needs to end with either 5 or 0. Using this as my base I know that all of the answers have to start and end with 5, as if it were 0 it wouldn't count. With this information all numbers start and end with 5.
ReplyDeleteMy strategy was to create 3 digit numbers that are multiples of 15, and ending with 5.
105 is the first 3 digit number ending in 5, making 501105. If I continue this pattern then the other results are as follows:
+Adding 30 to each 3 digit number is an easy way to interpret this.
+The 3 digit numbers end at 985.
501105 (already shown)
531135
561165
591195
522225
552255
582285
513315
543345
573375
504405
534435
564465
594495
525525
555555
585585
516615
546645
576675
517715
547745
577775
508805
538835
568865
598895
529925
559955
589985
To solve this question, we can narrow the answer down by stating a few rules.
ReplyDeleteThe number is a mirror image, so we can focus on the first 3 numbers
The number must be divisible by both 3 and 5
To be divisible by 5, it must end with either 0 or 5
It must start with whatever it ends in
It cannot start with 0, because then it won’t be a 6-digit number
It must start and end with 5
Now we know that the number is 5_ _ _ _5
Now, to find the second, third, fourth and fifth number:
The second and fifth number can be an digit from 0-9
Finally, the third and fourth digits must add up to create a multiple of 3.
Then, I used these rules, as well as trial and error to find all of the all the 6-digit palindromes divisible by 15.
501105
504405
507705
510015
513315
516615
519915
522225
525525
528825
531135
534435
537735
540045
543345
546645
549945
552255
555555
558855
561165
564465
567765
570075
573375
576675
579975
582285
585585
588885
591195
594495
597795
There are a total of 33 numbers.
All numbers start with 5. 15=3*5 which means the number ends with 0 or 5. Since your can't start with 0 it starts with 1.
ReplyDeleteI knew that for a number to be divisible by 15 it has to be divisible by 3 and 5. Therefore all of the numbers had to start with a 5 since a number can't start with a 0. Then just test all of the numbers that add to 6.
501105
504405
507705
510015
513315
516615
519915
522225
525525
528825
531135
534435
537735
540045
543345
546645
549945
552255
555555
558855
561165
564465
567765
570075
573375
576675
579975
582285
585585
588885
591195
594495
597795
POTW:
ReplyDeleteI am not sure whether or not I submitted my answer or not, but here it is:
There are about 31 different combinations.
The number has be divisible by 3 and 5, if it needs to be divisible by 15.
It needs to start with 5, as 0 wouldn't work:
Some examples include:
513315
510015
582285