Difference between revisions of "1991 OIM Problems/Problem 5"
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Now we try the possible combinations of <math>n</math> and <math>m</math>: | Now we try the possible combinations of <math>n</math> and <math>m</math>: | ||
+ | <math>\begin{cases} | ||
+ | m=0\text{; }n=0\text{; }P=2(0^2+0^2)+2(0+0)+1=1; & \text{YES}\\ | ||
+ | m=0\text{; }n=1\text{; }P=2(0^2+1^2)+2(0+1)+1=5; & \text{YES}\\ | ||
+ | m=0\text{; }n=2\text{; }P=2(0^2+2^2)+2(0+2)+1=13; & \text{YES}\\ | ||
+ | m=0\text{; }n=3\text{; }P=2(0^2+3^2)+2(0+3)+1=25; & \text{YES}\\ | ||
+ | m=0\text{; }n=4\text{; }P=2(0^2+4^2)+2(0+4)+1=41; & \text{YES}\\ | ||
+ | m=0\text{; }n=5\text{; }P=2(0^2+5^2)+2(0+5)+1=61; & \text{YES}\\ | ||
+ | m=0\text{; }n=6\text{; }P=2(0^2+6^2)+2(0+6)+1=85; & \text{YES}\\ | ||
+ | m=0\text{; }n=7\text{; }P=2(0^2+7^2)+2(0+7)+1=113; & \text{NO}\\ | ||
+ | m=1\text{; }n=1\text{; }P=2(1^2+1^2)+2(1+1)+1=9; & \text{YES}\\ | ||
+ | m=1\text{; }n=2\text{; }P=2(1^2+2^2)+2(1+2)+1=17; & \text{YES}\\ | ||
+ | m=1\text{; }n=3\text{; }P=2(1^2+3^2)+2(1+3)+1=29; & \text{YES}\\ | ||
+ | m=1\text{; }n=4\text{; }P=2(1^2+4^2)+2(1+4)+1=45; & \text{YES}\\ | ||
+ | m=1\text{; }n=5\text{; }P=2(1^2+5^2)+2(1+5)+1=65; & \text{YES}\\ | ||
+ | m=1\text{; }n=6\text{; }P=2(1^2+6^2)+2(1+6)+1=89; & \text{YES}\\ | ||
+ | m=1\text{; }n=7\text{; }P=2(1^2+7^2)+2(1+7)+1=117; & \text{NO}\\ | ||
+ | m=2\text{; }n=2\text{; }P=2(2^2+2^2)+2(2+2)+1=25; & \text{YES}\\ | ||
+ | m=2\text{; }n=3\text{; }P=2(2^2+3^2)+2(2+3)+1=37; & \text{YES}\\ | ||
+ | m=2\text{; }n=4\text{; }P=2(2^2+4^2)+2(2+4)+1=53; & \text{YES}\\ | ||
+ | m=2\text{; }n=5\text{; }P=2(2^2+5^2)+2(2+5)+1=73; & \text{YES}\\ | ||
+ | m=2\text{; }n=6\text{; }P=2(2^2+6^2)+2(2+6)+1=97; & \text{YES}\\ | ||
+ | m=2\text{; }n=7\text{; }P=2(2^2+7^2)+2(2+7)+1=125; & \text{NO}\\ | ||
+ | m=3\text{; }n=3\text{; }P=2(3^2+3^2)+2(3+3)+1=49; & \text{YES}\\ | ||
+ | m=3\text{; }n=4\text{; }P=2(3^2+4^2)+2(3+4)+1=65; & \text{YES}\\ | ||
+ | m=3\text{; }n=5\text{; }P=2(3^2+5^2)+2(3+5)+1=85; & \text{YES}\\ | ||
+ | m=3\text{; }n=6\text{; }P=2(3^2+6^2)+2(3+6)+1=109; & \text{NO}\\ | ||
+ | m=3\text{; }n=7\text{; }P=2(3^2+7^2)+2(3+7)+1=137; & \text{NO}\\ | ||
+ | m=4\text{; }n=4\text{; }P=2(4^2+4^2)+2(4+4)+1=81; & \text{YES}\\ | ||
+ | m=4\text{; }n=5\text{; }P=2(4^2+5^2)+2(4+5)+1=101; & \text{NO}\\ | ||
+ | m=4\text{; }n=6\text{; }P=2(4^2+6^2)+2(4+6)+1=125; & \text{NO}\\ | ||
+ | m=4\text{; }n=7\text{; }P=2(4^2+7^2)+2(4+7)+1=153; & \text{NO}\\ | ||
+ | m=5\text{; }n=5\text{; }P=2(5^2+5^2)+2(5+5)+1=121; & \text{NO}\\ | ||
+ | m=5\text{; }n=6\text{; }P=2(5^2+6^2)+2(5+6)+1=145; & \text{NO}\\ | ||
+ | m=5\text{; }n=7\text{; }P=2(5^2+7^2)+2(5+7)+1=173; & \text{NO}\\ | ||
+ | m=6\text{; }n=6\text{; }P=2(6^2+6^2)+2(6+6)+1=169; & \text{NO}\\ | ||
+ | m=6\text{; }n=7\text{; }P=2(6^2+7^2)+2(6+7)+1=197; & \text{NO}\\ | ||
+ | m=7\text{; }n=7\text{; }P=2(7^2+7^2)+2(7+7)+1=225; & \text{NO}\\ | ||
+ | \end{cases}</math> | ||
* Note. I actually competed at this event in Argentina when I was in High School representing Puerto Rico. I have no idea what I did on this one nor how many points they gave me. Probably close to zero on this one. | * Note. I actually competed at this event in Argentina when I was in High School representing Puerto Rico. I have no idea what I did on this one nor how many points they gave me. Probably close to zero on this one. |
Revision as of 20:28, 22 December 2023
Problem
Let . We will say that an integer is a value of if there exist integers and such that .
i. Determine how many elements of {1, 2, 3, ... ,100} are values of .
ii. Prove that the product of values of is a value of .
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
Part i.
Let , , be integers
, then solving for using the quadratic equation we have:
Let be an integer and . Therefore, Since , then , because
Since we can look at the combinations of with for non-negative values. So, we can use: to find the values of
Since , , then to get integers and , both expressions and need to be even. This happens when either and are both odd, or both even. Thus we will try both cases:
Case 1: Both and are even.
Let , where integers and with and
Now we try the possible combinations of and :
Case 2: Both and are odd.
Let , where integers and with and
Now we try the possible combinations of and :
- Note. I actually competed at this event in Argentina when I was in High School representing Puerto Rico. I have no idea what I did on this one nor how many points they gave me. Probably close to zero on this one.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.