Difference between revisions of "2024 AMC 10A Problems/Problem 1"
MRENTHUSIASM (talk | contribs) |
MRENTHUSIASM (talk | contribs) (→Solution 2 (Distributive Property)) |
||
| Line 12: | Line 12: | ||
== Solution 2 (Distributive Property) == | == Solution 2 (Distributive Property) == | ||
We have | We have | ||
| − | < | + | <cmath>\begin{align*} |
9901\cdot101-99\cdot10101 &= (10000-99)\cdot101-99\cdot(10000+101) \\ | 9901\cdot101-99\cdot10101 &= (10000-99)\cdot101-99\cdot(10000+101) \\ | ||
&= 10000\cdot101-99\cdot101-99\cdot10000-99\cdot101 \\ | &= 10000\cdot101-99\cdot101-99\cdot10000-99\cdot101 \\ | ||
| Line 18: | Line 18: | ||
&= 2\cdot10000-2\cdot9999 \\ | &= 2\cdot10000-2\cdot9999 \\ | ||
&= \boxed{\textbf{(A) }2}. | &= \boxed{\textbf{(A) }2}. | ||
| − | \end{align*} | + | \end{align*}</cmath> |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
Revision as of 15:19, 8 November 2024
Contents
Problem
What is the value of ![\[9901\cdot101-99\cdot10101?\]](http://latex.artofproblemsolving.com/1/3/4/134a62dd445139bc0163af4dc4a59a4de7086e6e.png)
Solution 1 (Direct Computation)
The likely fastest method will be direct computation. 
evaluates to 
and 
evaluates to 
. The difference is 
Solution by juwushu.
Solution 2 (Distributive Property)
We have
~MRENTHUSIASM
See also
| 2024 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
