https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Christina26&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T17:00:58ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10A_Problems/Problem_15&diff=802712002 AMC 10A Problems/Problem 152016-09-12T01:11:20Z<p>Christina26: /* Solution */</p>
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<div>==Problem==<br />
Using the digits 1, 2, 3, 4, 5, 6, 7, and 9, form 4 two-digit prime numbers, using each digit only once. What is the sum of the 4 prime numbers?<br />
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<math>\text{(A)}\ 150 \qquad \text{(B)}\ 160 \qquad \text{(C)}\ 170 \qquad \text{(D)}\ 180 \qquad \text{(E)}\ 190</math><br />
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==Solution==<br />
Only odd numbers can finish a two-digit prime number, and a two-digit number ending in 5 is divisible by 5 and thus composite, hence our answer is <math>20 + 40 + 50 + 60 + 1 + 3 + 7 + 9 = 190\Rightarrow\boxed{(E)= 190}</math>.<br />
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(Note that we did not need to actually construct the primes. If we had to, one way to match the tens and ones digits to form four primes is <math>23</math>, <math>41</math>, <math>59</math>, and <math>67</math>.)<br />
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(you could also guess the numbers, if you have a lot of spare time)<br />
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==See Also==<br />
{{AMC10 box|year=2002|ab=A|num-b=14|num-a=16}}<br />
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[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Christina26https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10A_Problems/Problem_15&diff=802702002 AMC 10A Problems/Problem 152016-09-12T00:58:19Z<p>Christina26: /* Solution */</p>
<hr />
<div>==Problem==<br />
Using the digits 1, 2, 3, 4, 5, 6, 7, and 9, form 4 two-digit prime numbers, using each digit only once. What is the sum of the 4 prime numbers?<br />
<br />
<math>\text{(A)}\ 150 \qquad \text{(B)}\ 160 \qquad \text{(C)}\ 170 \qquad \text{(D)}\ 180 \qquad \text{(E)}\ 190</math><br />
<br />
==Solution==<br />
Only odd numbers can finish a two-digit prime number, and a two-digit number ending in 5 is divisible by 5 and thus composite, hence our answer is <math>20 + 40 + 50 + 60 + 1 + 3 + 7 + 9 = 190\Rightarrow\boxed{(E)= 190}</math>.<br />
<br />
(Note that we did not need to actually construct the primes. If we had to, one way to match the tens and ones digits to form four primes is <math>23</math>, <math>41</math>, <math>59</math>, and <math>67</math>.)<br />
<br />
==See Also==<br />
{{AMC10 box|year=2002|ab=A|num-b=14|num-a=16}}<br />
<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Christina26https://artofproblemsolving.com/wiki/index.php?title=Simon%27s_Favorite_Factoring_Trick&diff=80248Simon's Favorite Factoring Trick2016-09-10T21:07:47Z<p>Christina26: /* The General Statement */</p>
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<div><br />
==About==<br />
'''Dr. Simon's Favorite Factoring Trick''' (abbreviated '''SFFT''') is a special factorization first popularized by [[AoPS]] user [[user:ComplexZeta | Simon Rubinstein-Salzedo]].<br />
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==The General Statement==<br />
The general statement of SFFT is: <math>{xy}+{xk}+{yj}+{jk}=(x+j)(y+k)</math>. Two special common cases are: <math>xy + x + y + 1 = (x+1)(y+1)</math> and <math>xy - x - y +1 = (x-1)(y-1)</math>.<br />
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The act of adding <math>{jk}</math> to <math>{xy}+{xk}+{yj}</math> in order to be able to factor it could be called "completing the rectangle" in analogy to the more familiar "completing the square."<br />
so funny!<br />
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== Applications ==<br />
This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually <math>x</math> and <math>y</math> are variables and <math>j,k</math> are known constants. Also, it is typically necessary to add the <math>jk</math> term to both sides to perform the factorization.<br />
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== Problems ==<br />
===Introductory===<br />
*Two different [[prime number]]s between <math>4</math> and <math>18</math> are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?<br />
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<math> \mathrm{(A) \ 21 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 } </math><br />
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([[2000 AMC 12/Problem 6|Source]])<br />
===Intermediate===<br />
*<math>m, n</math> are integers such that <math>m^2 + 3m^2n^2 = 30n^2 + 517</math>. Find <math>3m^2n^2</math>.<br />
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([[1987 AIME Problems/Problem 5|Source]])<br />
===Olympiad===<br />
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*The integer <math>N</math> is positive. There are exactly 2 pairs <math>(x, y)</math> of positive integers satisfying:<br />
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<cmath>\frac 1x +\frac 1y = \frac 1N</cmath><br />
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Prove that <math>N</math> is a perfect square. (British Mathematical Olympiad Round 2, 2005)<br />
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== See Also ==<br />
* [[Algebra]]<br />
* [[Factoring]]<br />
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[[Category:Elementary algebra]]<br />
[[Category:Theorems]]</div>Christina26https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_3&diff=800852013 AMC 8 Problems/Problem 32016-08-27T15:27:19Z<p>Christina26: /* Problem */</p>
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<div>==Problem==<br />
What is the value of <math>4 \cdot (-1+2-3+4-5+6-7+\cdots+1000)</math>? hihihihi<br />
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<math>\textbf{(A)}\ -10 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 500 \qquad \textbf{(E)}\ 2000</math><br />
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==Solution==<br />
Notice that we can pair up every two numbers to make a sum of 1:<br />
<cmath> \begin{eqnarray*}(-1 + 2 - 3 + 4 - \cdots + 1000) &=& ((-1 + 2) + (-3 + 4) + \cdots + (-999 + 1000)) \\ &=& (1 + 1 + \cdots + 1) \\ &=& 500\end{eqnarray*}</cmath><br />
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Therefore, the answer is <math>4 \cdot 500= \boxed{\textbf{(E)}\ 2000}</math>.<br />
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==See Also==<br />
{{AMC8 box|year=2013|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Christina26