Difference between revisions of "1969 Canadian MO Problems/Problem 1"
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(The denominator is irrelevant since it never equals zero) | (The denominator is irrelevant since it never equals zero) | ||
− | From <math>a_1/b_1= | + | From <math>a_1/b_1=a_2/b_2,</math> <math>a_1^nb_2^n=a_2^nb_1^n.</math> Similarly, <math>a_1^nb_3^n=a_3^nb_1^n</math> from <math>a_1/b_1=a_3/b_3.</math> |
Hence, <math>a_2^nb_1^n-a_1^nb_2^n=a_3^nb_1^n-a_1^nb_3^n=0</math> and our proof is complete. | Hence, <math>a_2^nb_1^n-a_1^nb_2^n=a_3^nb_1^n-a_1^nb_3^n=0</math> and our proof is complete. | ||
{{Old CanadaMO box|before=First question|num-a=2|year=1969}} | {{Old CanadaMO box|before=First question|num-a=2|year=1969}} |
Latest revision as of 15:06, 18 October 2015
Problem
Show that if and are not all zero, then for every positive integer
Solution
Instead of proving the two expressions equal, we prove that their difference equals zero.
Subtracting the LHS from the RHS,
Finding a common denominator, the numerator becomes (The denominator is irrelevant since it never equals zero)
From Similarly, from
Hence, and our proof is complete.
1969 Canadian MO (Problems) | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 2 |