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Hello everyone! I think that the number should be 59295. I started off by laying some rules down. Multiples of 15 always have either a 0 or 5, so I put the last and first number down as 5 (5 is obviously larger than 0), so I got: 5_ _ _5 If a follow the rules of adding all the digits, then so far, I have 10 (5+5). Because I only had 3 numbers left, I basically put the second and fourth blanks as 9, because 9 is the largest single digit number. Altogether, I had 28, and now had: 59_95 I only had one blank left, meaning that this last number added with all the other digits would make a number divisible by 3. the next number closest to 28 is 30, and 30-28 is 2, so naturally, I filled in the middle blank with 2.
With the number 59295, I divided it by 15, and got the number 3953, a whole number And there you have it! The highest 5 digit palindromic number divisible by 15 should be 59295.
Care to explain to others why you stated 'multiples of 15' always have to either 0 or 5, but then tested out also using the rule for multiple of 3s?
Also, in this part "I only had one blank left, meaning that this last number added with all the other digits would make a number divisible by 3. the next number closest to 28 is 30, and 30-28 is 2, so naturally, I filled in the middle blank with 2" I agree with your assertion EXCEPT the part when you stated "the next closest to 28"... if we are looking for the 'greatest' possible number, do we care for which is closest? (have you tried '36' instead of '30'?
We knew that the number we had to get needed to be divided exactly by 15, and that the sum of its digits would have to be divisible by 15, so we started off with a multiple of 15, in this case 45.
First we tried making all five digits add up to 45, so _ _ _ _ _ = 45.
We realized that one of the set of numbers could be 9, 9 ,9 ,9 ,9 since they all added up to 45 (meaning it would be divisible by 15), and it was a palindromic number.
* 9, 9, 9, 9, 9 backwards and forwards is the same
Since palindromic number never specified to be repeating numbers, I guess we can do it. :D
I came across this possibility as well, and although it is divisible by 3, it's not the same with 15. Although my answer was a bit off as well, I hope you found this comment useful! C:
in your first paragraph, you followed the rules that is used to determined whether 3 is a factor or not... the question however, calls for multiples of 15....what is the relationship between 3 and 15? consider it... =)
I agree that the last digit has to be either 5, or 0. 0 is out for obvious reason; however, there are still three digits to play with. Care to shed some lights on how you determine the middle three digits? there's no need for an elaborated response; but some reasoning would be helpful to us all~
Emily, do you see why it might be a third digit 8, and not a 2?
Tony, I do agree with Mr. H that you could use some explanation of your mathematical reasoning. After all, you DID get the correct answer. How about you let us in on how you got it? I know one way to solve this question. I begin with what it CAN'T be, then I worked towards what EACH of the five-digit needs to be in order to satisfy the parameters of the question.
We are looking for a fi ve-digit number of the form abcba. For a number to be divisible by 15, it must be divisible by both 3 and 5. To be divisible by 5, a number must end in 0 or 5. If the required number ends in 0, it must also begin with 0 in order to be a palindrome. But the number 0bcb0 is not a fi ve-digit number. Therefore, the number cannot end in a 0 and hence must start and end with a 5. The required number looks like 5bcb5. For a number to be divisible by 3, the sum of its digits must be divisible by 3. Since we want the largest possible number, let b = 9 in 5bcb5. We must fi nd the largest value of c so that 59c95 is divisible by 3. The sum of the digits is 5 + 9 + c + 9 + 5 = c + 28 and c can take on any integer value from 0 to 9. It follows that c + 28 can take on integer values from 28 to 37. The largest number in this range divisible by 3 is 36 so c + 28 = 36 and c = 8. The largest five-digit palindromic number exactly divisible by 15 is 59 895. (As an aside, the smallest fi ve-digit palindromic number exactly divisible by 15 is 50 205. And there are only 33 five-digit palindromic numbers exactly divisible by 15)
Hello everyone!
ReplyDeleteI think that the number should be 59295.
I started off by laying some rules down. Multiples of 15 always have either a 0 or 5, so I put the last and first number down as 5 (5 is obviously larger than 0), so I got:
5_ _ _5
If a follow the rules of adding all the digits, then so far, I have 10 (5+5).
Because I only had 3 numbers left, I basically put the second and fourth blanks as 9, because 9 is the largest single digit number. Altogether, I had 28, and now had:
59_95
I only had one blank left, meaning that this last number added with all the other digits would make a number divisible by 3. the next number closest to 28 is 30, and 30-28 is 2, so naturally, I filled in the middle blank with 2.
With the number 59295, I divided it by 15, and got the number 3953, a whole number
And there you have it! The highest 5 digit palindromic number divisible by 15 should be 59295.
Emily! Terrific explanation! Thank you.
DeleteCare to explain to others why you stated 'multiples of 15' always have to either 0 or 5, but then tested out also using the rule for multiple of 3s?
Also, in this part "I only had one blank left, meaning that this last number added with all the other digits would make a number divisible by 3. the next number closest to 28 is 30, and 30-28 is 2, so naturally, I filled in the middle blank with 2" I agree with your assertion EXCEPT the part when you stated "the next closest to 28"... if we are looking for the 'greatest' possible number, do we care for which is closest? (have you tried '36' instead of '30'?
=)
cheers,
Mr. Huang
Hello.
ReplyDeleteThe answer that my brother and I got was 99999.
We knew that the number we had to get needed to be divided exactly by 15, and that the sum of its digits would have to be divisible by 15, so we started off with a multiple of 15, in this case 45.
First we tried making all five digits add up to 45, so _ _ _ _ _ = 45.
We realized that one of the set of numbers could be 9, 9 ,9 ,9 ,9 since they all added up to 45 (meaning it would be divisible by 15), and it was a palindromic number.
* 9, 9, 9, 9, 9 backwards and forwards is the same
Since palindromic number never specified to be repeating numbers, I guess we can do it. :D
- Rachel and Jonathan
I came across this possibility as well, and although it is divisible by 3, it's not the same with 15.
DeleteAlthough my answer was a bit off as well, I hope you found this comment useful! C:
Sorry my mistake, I made a comment on my sister's account -.- And for the second time as well!
DeleteWell that was embarrassing.....
hi Rachel and Jon, thanks for your explanation!
Deletein your first paragraph, you followed the rules that is used to determined whether 3 is a factor or not... the question however, calls for multiples of 15....what is the relationship between 3 and 15? consider it... =)
cheers,
Mr. H
The answer is 59895
ReplyDeletebecause the first digit number has to be a 5
there are limmited possibilities after ou find that
hey Tony
DeleteI agree that the last digit has to be either 5, or 0. 0 is out for obvious reason; however, there are still three digits to play with. Care to shed some lights on how you determine the middle three digits? there's no need for an elaborated response; but some reasoning would be helpful to us all~
cheers,
Mr. H.
hi
ReplyDeleteEmily, do you see why it might be a third digit 8, and not a 2?
ReplyDeleteTony, I do agree with Mr. H that you could use some explanation of your mathematical reasoning. After all, you DID get the correct answer. How about you let us in on how you got it?
I know one way to solve this question. I begin with what it CAN'T be, then I worked towards what EACH of the five-digit needs to be in order to satisfy the parameters of the question.
We are looking for a fi ve-digit number of the form abcba.
For a number to be divisible by 15, it must be divisible by both 3 and 5.
To be divisible by 5, a number must end in 0 or 5. If the required number ends
in 0, it must also begin with 0 in order to be a palindrome. But the number
0bcb0 is not a fi ve-digit number. Therefore, the number cannot end in a 0 and
hence must start and end with a 5. The required number looks like 5bcb5.
For a number to be divisible by 3, the sum of its digits must be divisible by 3.
Since we want the largest possible number, let b = 9 in 5bcb5. We must fi nd
the largest value of c so that 59c95 is divisible by 3. The sum of the digits is
5 + 9 + c + 9 + 5 = c + 28 and c can take on any integer value from 0 to 9. It
follows that c + 28 can take on integer values from 28 to 37. The largest
number in this range divisible by 3 is 36 so c + 28 = 36 and c = 8.
The largest five-digit palindromic number exactly divisible by 15 is 59 895.
(As an aside, the smallest fi ve-digit palindromic number exactly divisible by 15
is 50 205. And there are only 33 five-digit palindromic numbers exactly divisible
by 15)