This blog is the online extension of our intermediate classrooms. Our goal is to enhance and document our learning experience throughout the school year, and share this journey with teachers, parents and students. We welcome your constructive feedback, and we look forward to learning with you!
The factorial question was certainly advanced for Intermediate, but sometimes it's nice to get a challenge! Please see the answer and hot to obtain it below (POTW #32 follows after that):
Actually, there is a correction to the question above. The number only satisfies the FIRST condition. It DOES NOT satisfy the third condition (because the sum of the digits is 54 NOT 45. Happy solving!
Started by fining patterns in the multiples of forty five and soon realized that they either end in 5 or 0. 0 wont work because then the first digit has to be zero and thats not a 7 digit number. So on both ends theres a five, meaning that so far, they add up to 10.
Then i thought the middle number should be an odd number to make the 2nd and 3rd digit, equal to the 5th and 6th. I placed 7 in the middle, and 6, and 8 in their proper spots. Basically i got the number 5687865.
I used division and found that this number was infact dividable by 45, answer being 126,397.
So 5687865 is a number that matches all the criteria.
Sorry for the poor explanation but this would be easier to expalin if there was a supporting image.
Using the power of guess and check, knowing that a number can only be divisible by 45 if it ends with 5 (meaning the number starts with 5), and knowing that numbers in the palindrome that are found twice need to have a sum between 36-45 (for the middle number to be a digit from 0-9), I have these numbers that meet the criteria: 567976, 5769675, 5793975, 5973795, 5891985, 5981895
To find all answers, I found several ways to get the sum of 45 by adding seven numbers from 0-9, 4 of which were different. 5 or 0 would have to be one of the digits because all multiples of 45 ends with 0 or 5. Using those numbers I made palindromes that are divisible by 45. Using my strategy i found these numbers, 5793975, 5769675, 5973795, 5891985.
After narrowing down what numbers could go where these are the possibilies: ( must start with 5 and end with 5 due being a mutiple of 45, and middle number must be and odd number so that the tens digit equals the hundred thousands digit and so that the hundreds digit equals the ten thousands digits) 5687865 5769675 5793975 5891985 5981895
Actually, there is a correction to the question above. The number only satisfies the FIRST condition. It DOES NOT satisfy the third condition (because the sum of the digits is 54 NOT 45. Happy solving!
ReplyDeleteStarted by fining patterns in the multiples of forty five and soon realized that they either end in 5 or 0. 0 wont work because then the first digit has to be zero and thats not a 7 digit number.
ReplyDeleteSo on both ends theres a five, meaning that so far, they add up to 10.
Then i thought the middle number should be an odd number to make the 2nd and 3rd digit, equal to the 5th and 6th. I placed 7 in the middle, and 6, and 8 in their proper spots. Basically i got the number 5687865.
I used division and found that this number was infact dividable by 45, answer being 126,397.
So 5687865 is a number that matches all the criteria.
Sorry for the poor explanation but this would be easier to expalin if there was a supporting image.
Using the power of guess and check, knowing that a number can only be divisible by 45 if it ends with 5 (meaning the number starts with 5), and knowing that numbers in the palindrome that are found twice need to have a sum between 36-45 (for the middle number to be a digit from 0-9), I have these numbers that meet the criteria:
ReplyDelete567976, 5769675, 5793975, 5973795, 5891985, 5981895
Sorry, first number I wrote should have been 5679765
DeleteI did mine hard copy
ReplyDeleteTo find all answers, I found several ways to get the sum of 45 by adding seven numbers from 0-9, 4 of which were different. 5 or 0 would have to be one of the digits because all multiples of 45 ends with 0 or 5. Using those numbers I made palindromes that are divisible by 45. Using my strategy i found these numbers, 5793975, 5769675, 5973795, 5891985.
ReplyDeleteAfter narrowing down what numbers could go where these are the possibilies: ( must start with 5 and end with 5 due being a mutiple of 45, and middle number must be and odd number so that the tens digit equals the hundred thousands digit and so that the hundreds digit equals the ten thousands digits)
ReplyDelete5687865
5769675
5793975
5891985
5981895