Difference between revisions of "1985 AJHSME Problems/Problem 1"
Scjh999999 (talk | contribs) m (→Solution 1) |
Scjh999999 (talk | contribs) m (→Solution 1) |
||
Line 8: | Line 8: | ||
==Solution 1== | ==Solution 1== | ||
− | By the [[associative property]], we can rearrange the numbers in the numerator and the denominator. [katex display=true]\frac{3}{3}\cdot \frac{5}{5}\cdot\frac{7}{7}\cdot\frac{9}{9}\cdot\frac{11}{11}=1\cdot1\cdot1\cdot1\cdot1=\boxed{\text{(A)} | + | By the [[associative property]], we can rearrange the numbers in the numerator and the denominator. [katex display=true]\frac{3}{3}\cdot \frac{5}{5}\cdot\frac{7}{7}\cdot\frac{9}{9}\cdot\frac{11}{11}=1\cdot1\cdot1\cdot1\cdot1=\boxed{\text{(A)} 1}[/katex] |
==Solution 2== | ==Solution 2== |
Revision as of 14:14, 23 January 2024
Problem
Solution 1
By the associative property, we can rearrange the numbers in the numerator and the denominator.
Solution 2
Notice that the in the denominator of the first fraction cancels with the same term in the second fraction, and the s in the numerator and denominator of the second fraction cancel. Then the expression is equal to .
~ cxsmi
Video Solution by BoundlessBrain!
See Also
1985 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.