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| − | {{duplicate|[[2024 AMC 12A Problems/Problem 1|2024 AMC 12A #1]] and [[2024 AMC 10A Problems/Problem 1|2024 AMC 10A #1]]}}
| + | #redirect[[2024 AMC 10A Problems/Problem 1]] |
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| − | ==Problem==
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| − | What is the value of <math>9901 \cdot 101 - 99 \cdot 10101</math>?
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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>
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| − | == Solution 1 (Direct Computation) ==
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| − | The likely fastest method will be direct computation. <math>9901 \cdot 101</math> evaluates to <math>1{,}000{,}001</math> and <math>99 \cdot 10101</math> evaluates to <math>999{,}999</math>. The difference is <math>\boxed{\textbf{(A)}~2}</math>.
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| − | Solution by [[User:Juwushu|juwushu]].
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| − | == Solution 2 (Distributive Property) ==
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| − | We have
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| − | <cmath>\begin{align*}
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| − | 9901\cdot101-99\cdot10101 &= (10000-99)\cdot101-99\cdot(10000+101) \
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| − | &= 10000\cdot101-99\cdot101-99\cdot10000-99\cdot101 \
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| − | &= (10000\cdot101-99\cdot10000)-2\cdot(99\cdot101) \
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| − | &= 2\cdot10000-2\cdot9999 \
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| − | &= \boxed{\textbf{(A) }2}.
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| − | \end{align*}</cmath>
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| − | ~MRENTHUSIASM
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| − | | |
| − | == Solution 3 (Solution 1 but Distributive) ==
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| − | Note that <math>9901\cdot101=9901\cdot100+9901=990100+9901=1000001</math> and <math>99\cdot10101=100\cdot10101-10101=1010100-10101=999999</math>, therefore the answer is <math>1000001-999999=\boxed{\textbf{(A) }2}</math>.
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| − | ~Tacos_are_yummy_1
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| − | == Solution 4 (Modular Arithmetic) ==
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| − | Evaluating the given expression <math>\pmod{10}</math> yields <math>1-9\equiv 2 \pmod{10}</math>, so the answer is either <math>\textbf{(A)}</math> or <math>\textbf{(D)}</math>. Evaluating <math>\pmod{101}</math> yields <math>0-99\equiv 2\pmod{101}</math>. Because answer <math>\textbf{(D)}</math> is <math>202=2\cdot 101</math>, that cannot be the answer, so we choose choice <math>\boxed{\textbf{(A) }2}</math>.
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| − | == Solution 5 (Process of Elimination) ==
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| − | We simply look at the units digit of the problem we have (or take mod <math>10</math>)
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| − | <cmath>9901\cdot101-99\cdot10101 \equiv 1\cdot1 - 9\cdot1 = 2 \mod{10}.</cmath>
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| − | Since the only answer with <math>2</math> in the units digit is <math>\textbf{(A)}</math>, We can then continue if you are desperate to use guess and check or a actually valid method to find the answer is <math>\boxed{\textbf{(A) }2}</math>.
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| − | ~[[User:Mathkiddus|mathkiddus]]
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| − | == Solution 6 (Faster Distribution) ==
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| − | Observe that <math>9901=9900+1=99\cdot100+1</math> and <math>10101=10100+1=101\cdot100+1</math>
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| − | <cmath>\begin{align*}
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| − | \Rightarrow9901\cdot101-99\cdot10101 & = ((9900\cdot101)+(1\cdot101))-((99\cdot10100)+(99\cdot1)) \
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| − | &=(99\cdot100\cdot101)+101-(99\cdot100\cdot101)-99 \
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| − | &=101-99 \
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| − | &=\boxed{\textbf{(A) }2}.
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| − | \end{align*}</cmath>
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| − | ~laythe_enjoyer211
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| − | ==Solution 7 (Cubes)==
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| − | Let <math>x=100</math>. Then, we have
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| − | \begin{align*}
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| − | 101\cdot 9901=(x+1)\cdot (x^2-x+1)=x^3+1, \
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| − | 99\cdot 10101=(x-1)\cdot (x^2+x+1)=x^3-1.
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| − | \end{align*}
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| − | Then, the answer can be rewritten as <math>(x^3+1)-(x^3-1)= \boxed{\textbf{(A) }2}.</math>
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| − | ~erics118
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| − | ==Solution 8 (Super Fast)==
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| − | It's not hard to observe and express <math>9901</math> into <math>99\cdot100+1</math>, and <math>10101</math> into <math>101\cdot100+1</math>.
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| − | We then simplify the original expression into <math>(99\cdot100+1)\cdot101-99\cdot(101\cdot100+1)</math>, which could then be simplified into <math>99\cdot100\cdot101+101-99\cdot100\cdot101-99</math>, which we can get the answer of <math>101-99=\boxed{\textbf{(A) }2}</math>.
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| − | ~RULE101
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| − | == Video Solution (⚡️ 1 min solve ⚡️) ==
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| − | https://youtu.be/RODYXdpipdc
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| − | <i>~Education, the Study of Everything </i>
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| − | == Video Solution by Pi Academy ==
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| − | https://youtu.be/GPoTfGAf8bc?si=JYDhLVzfHUbXa3DW
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| − | == Video Solution by FrankTutor ==
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| − | https://www.youtube.com/watch?v=ez095SvW5xI
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| − | == Video Solution Daily Dose of Math ==
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| − | https://youtu.be/Z76bafQsqTc
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| − | ~Thesmartgreekmathdude
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| − | == Video Solution 1 by Power Solve ==
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| − | https://www.youtube.com/watch?v=j-37jvqzhrg
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| − | ==Video Solution by SpreadTheMathLove==
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| − | https://www.youtube.com/watch?v=6SQ74nt3ynw
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| − | ==Video Solution by Math from my desk ==
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| − | https://www.youtube.com/watch?v=n_G6wi1ulzY
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| − | ==See also==
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| − | {{AMC12 box|year=2024|ab=A|before=First Problem|num-a=2}}
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| − | {{AMC10 box|year=2024|ab=A|before=First Problem|num-a=2}}
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| − | {{MAA Notice}}
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