Difference between revisions of "2023 CEMC Cayley Problems/Problem 1"
(Created page with "==Problem== What is the value of <math>\frac{1}{1} + \frac{2}{2} + \frac{3}{3}</math>? <math> \text{ (A) }\ 1 \qquad\text{ (B) }\ 0 \qquad\text{ (C) }\ 3 \qquad\text{ (D) }\...") |
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<math>\frac{1 \times 6}{1 \times 6} + \frac{2 \times 3}{2 \times 3} + \frac{3 \times 2}{3 \times 2}</math> | <math>\frac{1 \times 6}{1 \times 6} + \frac{2 \times 3}{2 \times 3} + \frac{3 \times 2}{3 \times 2}</math> | ||
| − | <math>=\frac{6}{6} + \frac{6}{6} + \frac{6}{6} = \frac{18}{6} = 3</math> | + | <math>=\frac{6}{6} + \frac{6}{6} + \frac{6}{6} = \frac{18}{6} = \boxed {\textbf {(C) } 3}</math> |
~anabel.disher | ~anabel.disher | ||
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If <math>x \neq 0</math>, then <math>\frac{x}{x} = \frac{x \div x}{x \div x} = \frac{1}{1} = 1</math>. Thus, we have: | If <math>x \neq 0</math>, then <math>\frac{x}{x} = \frac{x \div x}{x \div x} = \frac{1}{1} = 1</math>. Thus, we have: | ||
| − | <math>1 + 1 + 1 = 3</math> | + | <math>1 + 1 + 1 = \boxed {\textbf {(C) } 3}</math> |
~anabel.disher | ~anabel.disher | ||
Latest revision as of 13:29, 11 September 2025
Problem
What is the value of 
?
Solution 1
Using the greatest common denominator, we have:
~anabel.disher
Solution 2
If 
, then 
. Thus, we have:
~anabel.disher
