Difference between revisions of "2024 AMC 10A Problems/Problem 1"

(Solution 3 (Quickest Way))
(Solution 3 (A way to get rid of some answer choices(Not recommended on the actual competition))
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~MRENTHUSIASM
 
~MRENTHUSIASM
== Solution 3 (A way to get rid of some answer choices(Not recommended on the actual competition) ==
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== Solution 3 (Process of Elimination) (Not recommended) ==
  
 
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)

Revision as of 15:31, 8 November 2024

Problem

What is the value of \[9901\cdot101-99\cdot10101?\]

$\textbf{(A)}~2\qquad\textbf{(B)}~20\qquad\textbf{(C)}~200\qquad\textbf{(D)}~202\qquad\textbf{(E)}~2020$

Solution 1 (Direct Computation)

The likely fastest method will be direct computation. $9901\cdot101$ evaluates to $1000001$ and $99\cdot10101$ evaluates to $999999$. The difference is $\boxed{\textbf{(A) }2}.$

Solution by juwushu.

Solution 2 (Distributive Property)

We have \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\cdot(99\cdot101) \\ &= 2\cdot10000-2\cdot9999 \\ &= \boxed{\textbf{(A) }2}. \end{align*} ~MRENTHUSIASM

Solution 3 (Process of Elimination) (Not recommended)

We simply look at the units digit of the problem we have(or take mod 10) \[9901\cdot101-99\cdot10101 \equiv 1\cdot1 - 9\cdot1 = 2 \mod{10}.\] Since the only answer with 2 in the units digit is $\textbf{(A)}$ or $\textbf{(D)}$ We can then continue if you are desperate to use guess and check or a actually valid method to find the answer is $\boxed{\textbf{(A)}}$ ~mathkiddus

See also

2024 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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All AMC 10 Problems and Solutions

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