Question 1209286: All numbers (N) that are written in three decimal places, divisible by 11, and when divided by 11, the quotient equals the sum of the squares of its three digits.
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
I think the phrasing "three decimal places" should be "three digits".
Also I'll assume we focus on whole numbers only.
If those assumptions are the case then we're looking through the set {100, 101, 102, ..., 998, 999}
100/11 = 9.09 approximately
999/11 = 90.82 approximately
11*9 = 99 is too small, but 11*10 = 110 fits.
11*91 = 1001 is too large, but 11*90 = 990 fits.
The smallest and largest multiple of 11 in this set would be 110 and 990.
We have reduced the set to {110, 121, 132, ..., 979, 990}
Which you can think of as {11*10, 11*11, 11*12, ..., 11*89, 11*90} where we multiply 11 by items inside {10, 11, 12, ..., 89, 90}
Each blue item is a quotient when dividing over 11.
Two examples would be 110/11 = 10 and 979/11 = 89
Then you can use trial-and-error to square the digits and add up the results
110 ---> 1^2+1^2+0^2 = 2 which doesn't match with the quotient 110/11 = 10
121 ---> 1^2+2^2+1^2 = 6 which doesn't match with the quotient 121/11 = 11
132 ---> 1^2+3^2+2^2 = 14 which doesn't match with the quotient 132/11 = 12
143 ---> 1^2+4^2+3^2 = 26 which doesn't match with the quotient 143/11 = 13
And so on.
This might take a while.
To speed things up I made a Python script to search through all the values.
The program found the following solutions: 550, 803
550 ---> 5^2+5^2+0^2 = 50 which matches with the quotient 550/11 = 50
803 ---> 8^2+0^2+3^2 = 73 which matches with the quotient 803/11 = 73
There might be a much more efficient approach.
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
See the complete mathematical reasoning/solution under this link
https://math.stackexchange.com/questions/1805667/determine-all-three-digit-numbers-n-having-the-property-that-n-is-divisible-by-1
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