Difference between revisions of "2024 AMC 10A Problems/Problem 1"
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We simply look at the units digit of the problem we have(or take mod 10) | We simply look at the units digit of the problem we have(or take mod 10) | ||
| − | <cmath>9901\cdot101-99\cdot10101 \equiv 1 | + | <cmath>9901\cdot101-99\cdot10101 \equiv 1\cdot1 - 9\cdot1 = 2 \mod{10}.</cmath> |
Since the only answer with 2 in the units digit is <math>\boxed{\textbf{(A)}}</math> we are done! | Since the only answer with 2 in the units digit is <math>\boxed{\textbf{(A)}}</math> we are done! | ||
~mathkiddus | ~mathkiddus | ||
Revision as of 15:28, 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
Solution 3 (Quickest Way)
We simply look at the units digit of the problem we have(or take mod 10)
![\[9901\cdot101-99\cdot10101 \equiv 1\cdot1 - 9\cdot1 = 2 \mod{10}.\]](http://latex.artofproblemsolving.com/c/8/e/c8ec864e059d92cfd87d3afbeeb0f6ced81dbe23.png)
Since the only answer with 2 in the units digit is 
we are done!
~mathkiddus
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. 
