Difference between revisions of "2024 AMC 10A Problems/Problem 1"
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<math>\textbf{(A)}~2\qquad\textbf{(B)}~20\qquad\textbf{(C)}~200\qquad\textbf{(D)}~202\qquad\textbf{(E)}~2020</math> | <math>\textbf{(A)}~2\qquad\textbf{(B)}~20\qquad\textbf{(C)}~200\qquad\textbf{(D)}~202\qquad\textbf{(E)}~2020</math> | ||
| − | == Solution 1 == | + | == Solution 1 (Direct Computation) == |
| − | The likely fastest method will be | + | The likely fastest method will be direct computation. <math>9901\cdot101</math> evaluates to <math>1000001</math> and <math>99\cdot10101</math> evaluates to <math>999999</math>. The difference is <math>\boxed{\textbf{(A) }2}.</math> |
Solution by [[User:Juwushu|juwushu]]. | Solution by [[User:Juwushu|juwushu]]. | ||
| + | |||
| + | == Solution 2 (Distributive Property) == | ||
| + | We have | ||
| + | <math></math>\begin{align*} | ||
| + | 9901\cdot101-99\cdot10101 &= (10000-99)\cdot101-99\cdot(10000+101) \\ | ||
| + | &= 10000\cdot101-99\cdot101-99\cdot10000-99\cdot101 \\ | ||
| + | &= (10000\cdot101-99\cdot10000)-2\cdot99\cdot101 \\ | ||
| + | &= 2\cdot10000-2\cdot9999 \\ | ||
| + | &= \boxed{\textbf{(A) }2}. | ||
| + | \end{align*} | ||
| + | ~MRENTHUSIASM | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=A|before=First Problem|num-a=2}} | {{AMC10 box|year=2024|ab=A|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
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 $$ (Error compiling LaTeX. Unknown error_msg)\begin{align*} 9901\cdot101-99\cdot10101 &= (10000-99)\cdot101-99\cdot(10000+101) \\ &= 10000\cdot101-99\cdot101-99\cdot10000-99\cdot101 \\ &= (10000\cdot101-99\cdot10000)-2\cdot99\cdot101 \\ &= 2\cdot10000-2\cdot9999 \\ &= \boxed{\textbf{(A) }2}. \end{align*} ~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. 
