Difference between revisions of "1991 OIM Problems/Problem 5"
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<math>\begin{cases} | <math>\begin{cases} | ||
− | m=0\text{; }n=0\text{; }P=(0^2+0^2) | + | m=0\text{; }n=0\text{; }P=2(0^2+0^2)=0; & \text{NO}\\ |
− | m=0\text{; }n=1\text{; }P=(0^2+1^2) | + | m=0\text{; }n=1\text{; }P=2(0^2+1^2)=2; & \text{YES}\\ |
− | m=0\text{; }n=2\text{; }P=(0^2+2^2) | + | m=0\text{; }n=2\text{; }P=2(0^2+2^2)=8; & \text{YES}\\ |
− | m=0\text{; }n=3\text{; }P=(0^2+3^2) | + | m=0\text{; }n=3\text{; }P=2(0^2+3^2)=18; & \text{YES}\\ |
− | m=0\text{; }n=4\text{; }P=(0^2+4^2) | + | m=0\text{; }n=4\text{; }P=2(0^2+4^2)=32; & \text{YES}\\ |
− | m=0\text{; }n=5\text{; }P=(0^2+5^2) | + | m=0\text{; }n=5\text{; }P=2(0^2+5^2)=50; & \text{YES}\\ |
− | m=0\text{; }n=6\text{; }P=(0^2+6^2) | + | m=0\text{; }n=6\text{; }P=2(0^2+6^2)=72; & \text{YES}\\ |
− | m=0\text{; }n=7\text{; }P=(0^2+7^2) | + | m=0\text{; }n=7\text{; }P=2(0^2+7^2)=98; & \text{YES}\\ |
− | m=1\text{; }n=1\text{; }P=(1^2+1^2) | + | m=1\text{; }n=1\text{; }P=2(1^2+1^2)=4; & \text{YES}\\ |
− | m=1\text{; }n=2\text{; }P=(1^2+2^2) | + | m=1\text{; }n=2\text{; }P=2(1^2+2^2)=10; & \text{YES}\\ |
− | m=1\text{; }n=3\text{; }P=(1^2+3^2) | + | m=1\text{; }n=3\text{; }P=2(1^2+3^2)=20; & \text{YES}\\ |
− | m=1\text{; }n=4\text{; }P=(1^2+4^2) | + | m=1\text{; }n=4\text{; }P=2(1^2+4^2)=34; & \text{YES}\\ |
− | m=1\text{; }n=5\text{; }P=(1^2+5^2) | + | m=1\text{; }n=5\text{; }P=2(1^2+5^2)=52; & \text{YES}\\ |
− | m=1\text{; }n=6\text{; }P=(1^2+6^2) | + | m=1\text{; }n=6\text{; }P=2(1^2+6^2)=74; & \text{YES}\\ |
− | m=1\text{; }n=7\text{; }P=(1^2+7^2) | + | m=1\text{; }n=7\text{; }P=2(1^2+7^2)=100; & \text{YES}\\ |
− | m=2\text{; }n=2\text{; }P=(2^2+2^2) | + | m=2\text{; }n=2\text{; }P=2(2^2+2^2)=16; & \text{YES}\\ |
− | m=2\text{; }n=3\text{; }P=(2^2+3^2) | + | m=2\text{; }n=3\text{; }P=2(2^2+3^2)=26; & \text{YES}\\ |
− | m=2\text{; }n=4\text{; }P=(2^2+4^2) | + | m=2\text{; }n=4\text{; }P=2(2^2+4^2)=40; & \text{YES}\\ |
− | m=2\text{; }n=5\text{; }P=(2^2+5^2) | + | m=2\text{; }n=5\text{; }P=2(2^2+5^2)=58; & \text{YES}\\ |
− | m=2\text{; }n=6\text{; }P=(2^2+6^2) | + | m=2\text{; }n=6\text{; }P=2(2^2+6^2)=80; & \text{YES}\\ |
− | m=2\text{; }n=7\text{; }P=(2^2+7^2) | + | m=2\text{; }n=7\text{; }P=2(2^2+7^2)=106; & \text{NO}\\ |
− | m=3\text{; }n=3\text{; }P=(3^2+3^2) | + | m=3\text{; }n=3\text{; }P=2(3^2+3^2)=36; & \text{YES}\\ |
− | m=3\text{; }n=4\text{; }P=(3^2+4^2) | + | m=3\text{; }n=4\text{; }P=2(3^2+4^2)=50; & \text{YES}\\ |
− | m=3\text{; }n=5\text{; }P=(3^2+5^2) | + | m=3\text{; }n=5\text{; }P=2(3^2+5^2)=68; & \text{YES}\\ |
− | m=3\text{; }n=6\text{; }P=(3^2+6^2) | + | m=3\text{; }n=6\text{; }P=2(3^2+6^2)=90; & \text{YES}\\ |
− | m=3\text{; }n=7\text{; }P=(3^2+7^2) | + | m=3\text{; }n=7\text{; }P=2(3^2+7^2)=116; & \text{NO}\\ |
− | m=4\text{; }n=4\text{; }P=(4^2+4^2) | + | m=4\text{; }n=4\text{; }P=2(4^2+4^2)=64; & \text{YES}\\ |
− | m=4\text{; }n=5\text{; }P=(4^2+5^2) | + | m=4\text{; }n=5\text{; }P=2(4^2+5^2)=82; & \text{YES}\\ |
− | m=4\text{; }n=6\text{; }P=(4^2+6^2) | + | m=4\text{; }n=6\text{; }P=2(4^2+6^2)=104; & \text{NO}\\ |
− | m=4\text{; }n=7\text{; }P=(4^2+7^2) | + | m=4\text{; }n=7\text{; }P=2(4^2+7^2)=130; & \text{NO}\\ |
− | m=5\text{; }n=5\text{; }P=(5^2+5^2) | + | m=5\text{; }n=5\text{; }P=2(5^2+5^2)=100; & \text{YES}\\ |
− | m=5\text{; }n=6\text{; }P=(5^2+6^2) | + | m=5\text{; }n=6\text{; }P=2(5^2+6^2)=122; & \text{NO}\\ |
− | m=5\text{; }n=7\text{; }P=(5^2+7^2) | + | m=5\text{; }n=7\text{; }P=2(5^2+7^2)=148; & \text{NO}\\ |
− | m=6\text{; }n=6\text{; }P=(6^2+6^2) | + | m=6\text{; }n=6\text{; }P=2(6^2+6^2)=144; & \text{NO}\\ |
− | m=6\text{; }n=7\text{; }P=(6^2+7^2) | + | m=6\text{; }n=7\text{; }P=2(6^2+7^2)=170; & \text{NO}\\ |
− | m=7\text{; }n=7\text{; }P=(7^2+7^2) | + | m=7\text{; }n=7\text{; }P=2(7^2+7^2)=196; & \text{NO} |
\end{cases}</math> | \end{cases}</math> | ||
Revision as of 20:23, 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 :
- 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.