Difference between revisions of "2024 AMC 10A Problems/Problem 1"
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~mathkiddus | ~mathkiddus | ||
== Solution 4 (Solution 1 but distrubutive) == | == Solution 4 (Solution 1 but distrubutive) == | ||
| − | Note that <math>9901\cdot101=9901\cdot100+9901=9901=990100+9901=1000001</math> and <math>99\cdot10101=100\cdot10101-10101=1010100-10101=999999, therefore the answer is < | + | Note that <math>9901\cdot101=9901\cdot100+9901=9901=990100+9901=1000001</math> and <math>99\cdot10101=100\cdot10101-10101=1010100-10101=999999</math>, therefore the answer is <math>1000001-999999=\boxed{\text{(A) }2}</math> |
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==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 15:45, 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 (Process of Elimination) (Not recommended)
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 
or 
We can then continue if you are desperate to use guess and check or a actually valid method to find the answer is 
~mathkiddus
Solution 4 (Solution 1 but distrubutive)
Note that 
and 
, therefore the answer is 
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. 
