Difference between revisions of "1985 AJHSME Problems/Problem 1"
Megaboy6679 (talk | contribs) (→Solution 1) |
|||
Line 7: | Line 7: | ||
==Solution 1== | ==Solution 1== | ||
− | Noticing that multiplying and dividing by the same number is the equivalent of multiplying (or dividing) by <math>1</math>, we can rearrange the numbers in the numerator and the denominator ([[ | + | Noticing that multiplying and dividing by the same number is the equivalent of multiplying (or dividing) by <math>1</math>, we can rearrange the numbers in the numerator and the denominator ([[Associative property| associative property of multiplication]]) so that it looks like <cmath>\frac{3}{3} \times \frac{5}{5} \times \frac{7}{7} \times \frac{9}{9} \times \frac{11}{11}.</cmath> |
Notice that each number is still there, and nothing has been changed - other than the order. | Notice that each number is still there, and nothing has been changed - other than the order. | ||
− | Finally, since all of the fractions are equal to one, we have | + | Finally, since all of the fractions are equal to one, we have $1\cdot1\cdot1\cdot1\cdot1=\boxed{\textbf{A} 1}. |
− | |||
− | |||
==Solution 2 (Brute force)== | ==Solution 2 (Brute force)== |
Revision as of 19:59, 1 February 2023
Problem
Solution 1
Noticing that multiplying and dividing by the same number is the equivalent of multiplying (or dividing) by , we can rearrange the numbers in the numerator and the denominator ( associative property of multiplication) so that it looks like
Notice that each number is still there, and nothing has been changed - other than the order.
Finally, since all of the fractions are equal to one, we have $1\cdot1\cdot1\cdot1\cdot1=\boxed{\textbf{A} 1}.
Solution 2 (Brute force)
If you want to multiply it out, then it would be
That would be which is 1. Therefore, the answer is .
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.