Difference between revisions of "1960 IMO Problems/Problem 1"
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==Solution== | ==Solution== | ||
| + | {{non-rigorous}} | ||
| − | {{ | + | Define a '''ten''' to be all ten positive integers which begin with a fixed tens digit. |
| + | |||
| + | We can make a systematic approach to this: | ||
| + | |||
| + | |||
| + | By inspection, <math>\dfrac{N}{11}</math> must be between 10 and 90 inclusive. That gives us 8 tens to check, and 90 as well. | ||
| + | |||
| + | For a given ten, the sum of the squares of the digits of <math>N</math> increases faster than <math>\dfrac{N}{11}</math>, so we can have at most one number in every ten that works. | ||
| + | |||
| + | We check the first ten: | ||
| + | |||
| + | <math>11*11=121</math> | ||
| + | |||
| + | <math>1^2+2^2+1^2=4</math> | ||
| + | |||
| + | <math>12*11=132</math> | ||
| + | |||
| + | <math>1^2+3^2+2^2=14</math> | ||
| + | |||
| + | 11 is too small and 12 is too large, so all numbers below 11 will be too small and all numbers above 12 will be too large, so no numbers in the first ten work. | ||
| + | |||
| + | We try the second ten: | ||
| + | |||
| + | <math>21*11=231</math> | ||
| + | |||
| + | <math>2^2+3^2+1^2=14</math> | ||
| + | |||
| + | <math>22*11=242</math> | ||
| + | |||
| + | <math>2^2+4^2+2^2=24</math> | ||
| + | |||
| + | Therefore, no numbers n the second ten work. | ||
| + | |||
| + | We continue, to find out that 50 is the only one that works. | ||
| + | |||
| + | <math>N=50*11=550</math>, so there s only one <math>N</math> that works. | ||
| + | |||
| + | {{alternate solutions}} | ||
==See Also== | ==See Also== | ||
{{IMO box|year=1960|before=First Question|num-a=2}} | {{IMO box|year=1960|before=First Question|num-a=2}} | ||
Revision as of 11:19, 28 October 2007
Problem
Determine all three-digit numbers 
having the property that is divisible by 11, and 
is equal to the sum of the squares of the digits of .
Solution
Define a ten to be all ten positive integers which begin with a fixed tens digit.
We can make a systematic approach to this:
By inspection, must be between 10 and 90 inclusive. That gives us 8 tens to check, and 90 as well.
For a given ten, the sum of the squares of the digits of increases faster than , so we can have at most one number in every ten that works.
We check the first ten:
11 is too small and 12 is too large, so all numbers below 11 will be too small and all numbers above 12 will be too large, so no numbers in the first ten work.
We try the second ten:
Therefore, no numbers n the second ten work.
We continue, to find out that 50 is the only one that works.
, so there s only one that works.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
| 1960 IMO (Problems) • Resources | ||
| Preceded by First Question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
| All IMO Problems and Solutions | ||
