https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Peashooter6969&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T14:05:53ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1985_AJHSME_Problems/Problem_2&diff=943311985 AJHSME Problems/Problem 22018-05-07T01:04:31Z<p>Peashooter6969: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
<br />
<math>90+91+92+93+94+95+96+97+98+99=</math><br />
<br />
<br />
<math>\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045</math><br />
<br />
==Solution==<br />
<br />
===Solution 1===<br />
One possibility is to simply add them. However, this can be time-consuming, and there are other ways to solve this problem. <br />
We find a simpler problem in this problem, and simplify -> <math>90 + 91 + ... + 98 + 99 = 90 \times 10 + 1 + 2 + 3 + ... + 8 + 9</math><br />
<br />
We know <math>90 \times 10</math>, that's easy - <math>900</math>. So how do we find <math>1 + 2 + ... + 8 + 9</math>?<br />
<br />
We rearrange the numbers to make <math>(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5</math>. You might have noticed that each of the terms we put next to each other add up to 10, which makes for easy adding. <math>4 \times 10 + 5 = 45</math>. Adding that on to 900 makes 945.<br />
<br />
945 is <math>\boxed{\text{B}}</math><br />
<br />
===Solution 2===<br />
Instead of breaking the sum and then rearranging, we can start by rearranging:<br />
<cmath>\begin{align*}<br />
90+91+92+\cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\<br />
&= 189+189+189+189+189 \\<br />
&= 945\rightarrow \boxed{\text{B}} <br />
\end{align*}</cmath><br />
<br />
===Solution 3===<br />
<br />
We can use the formula for finite arithmetic sequences.<br />
<br />
It is <math>\frac{n}{2}\times</math> (<math>a_1+a_n</math>) where <math>n</math> is the number of terms in the sequence, <math>a_1</math> is the first term and <math>a_n</math> is the last term.<br />
<br />
Applying it here:<br />
<br />
<math>\frac{10}{2} \times (90+99) = 945 \rightarrow \boxed{B}</math><br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1985|num-b=1|num-a=3}}<br />
<br />
<br />
{{MAA Notice}}<br />
[[Category:Introductory Algebra Problems]]</div>Peashooter6969https://artofproblemsolving.com/wiki/index.php?title=1985_AJHSME_Problems/Problem_2&diff=943301985 AJHSME Problems/Problem 22018-05-07T00:59:44Z<p>Peashooter6969: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
<br />
<math>90+91+92+93+94+95+96+97+98+99=</math><br />
<br />
<br />
<math>\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045</math><br />
<br />
==Solution==<br />
<br />
===Solution 1===<br />
One possibility is to simply add them. However, this can be time-consuming, and there are other ways to solve this problem. <br />
We find a simpler problem in this problem, and simplify -> <math>90 + 91 + ... + 98 + 99 = 90 \times 10 + 1 + 2 + 3 + ... + 8 + 9</math><br />
<br />
We know <math>90 \times 10</math>, that's easy - <math>900</math>. So how do we find <math>1 + 2 + ... + 8 + 9</math>?<br />
<br />
We rearrange the numbers to make <math>(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5</math>. You might have noticed that each of the terms we put next to each other add up to 10, which makes for easy adding. <math>4 \times 10 + 5 = 45</math>. Adding that on to 900 makes 945.<br />
<br />
945 is <math>\boxed{\text{B}}.</math><br />
<br />
===Solution 2===<br />
Instead of breaking the sum and then rearranging, we can start by rearranging:<br />
<cmath>\begin{align*}<br />
90+91+92+\cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\<br />
&= 189+189+189+189+189 \\<br />
&= 945\rightarrow \boxed{\text{B}} <br />
\end{align*}</cmath><br />
<br />
===Solution 3===<br />
<br />
We can use the formula for finite arithmetic sequences.<br />
<br />
It is <math>\frac{n}{2}\times</math> (<math>a_1+a_n</math>) where <math>n</math> is the number of terms in the sequence, <math>a_1</math> is the first term and <math>a_n</math> is the last term.<br />
<br />
Applying it here:<br />
<br />
<math>\frac{10}{2} \times (90+99) = 945 \rightarrow \boxed{B}</math><br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1985|num-b=1|num-a=3}}<br />
<br />
<br />
{{MAA Notice}}<br />
[[Category:Introductory Algebra Problems]]</div>Peashooter6969https://artofproblemsolving.com/wiki/index.php?title=1985_AJHSME_Problems/Problem_2&diff=943291985 AJHSME Problems/Problem 22018-05-07T00:59:18Z<p>Peashooter6969: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
<br />
<math>90+91+92+93+94+95+96+97+98+99=</math><br />
<br />
<br />
<math>\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045</math><br />
<br />
==Solution==<br />
<br />
===Solution 1===<br />
One possibility is to simply add them. However, this can be time-consuming, and there are other ways to solve this problem. <br />
We find a simpler problem in this problem, and simplify -> <math>90 + 91 + ... + 98 + 99 = 90 \times 10 + 1 + 2 + 3 + ... + 8 + 9</math><br />
<br />
We know <math>90 \times 10</math>, that's easy - <math>900</math>. So how do we find <math>1 + 2 + ... + 8 + 9</math>?<br />
<br />
We rearrange the numbers to make <math>(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5</math>. You might have noticed that each of the terms we put next to each other add up to 10, which makes for easy adding. <math>4 \times 10 + 5 = 45</math>. Adding that on to 900 makes 945.<br />
<br />
945 is <math>\boxed{\text{B}}.</math><br />
<br />
===Solution 2===<br />
Instead of breaking the sum and then rearranging, we can start by rearranging:<br />
<cmath>\begin{align*}<br />
90+91+92+\cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\<br />
&= 189+189+189+189+189 \\<br />
&= 945\rightarrow \boxed{\text{B}} <br />
\end{align*}</cmath><br />
<br />
===Solution 3===<br />
<br />
We can use the formula for finite arithmetic sequences.<br />
<br />
It is <math>\frac{n}{2}\times</math> (<math>a_1+a_n</math>) where <math>n</math> is the number of terms in the sequence, <math>a_1</math> is the first term and <math>a_n</math> is the last term<br />
<br />
Applying it here:<br />
<br />
<math>\frac{10}{2} \times (90+99) = 945 \rightarrow \boxed{B}</math><br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1985|num-b=1|num-a=3}}<br />
<br />
<br />
{{MAA Notice}}<br />
[[Category:Introductory Algebra Problems]]</div>Peashooter6969https://artofproblemsolving.com/wiki/index.php?title=1985_AJHSME_Problems/Problem_2&diff=943281985 AJHSME Problems/Problem 22018-05-07T00:50:41Z<p>Peashooter6969: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
<br />
<math>90+91+92+93+94+95+96+97+98+99=</math><br />
<br />
<br />
<math>\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045</math><br />
<br />
==Solution==<br />
<br />
===Solution 1===<br />
One possibility is to simply add them. However, this can be time-consuming, and there are other ways to solve this problem. <br />
We find a simpler problem in this problem, and simplify -> <math>90 + 91 + ... + 98 + 99 = 90 \times 10 + 1 + 2 + 3 + ... + 8 + 9</math><br />
<br />
We know <math>90 \times 10</math>, that's easy - <math>900</math>. So how do we find <math>1 + 2 + ... + 8 + 9</math>?<br />
<br />
We rearrange the numbers to make <math>(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5</math>. You might have noticed that each of the terms we put next to each other add up to 10, which makes for easy adding. <math>4 \times 10 + 5 = 45</math>. Adding that on to 900 makes 945.<br />
<br />
945 is <math>\boxed{\text{B}}.</math><br />
<br />
===Solution 2===<br />
Instead of breaking the sum and then rearranging, we can start by rearranging:<br />
<cmath>\begin{align*}<br />
90+91+92+\cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\<br />
&= 189+189+189+189+189 \\<br />
&= 945\rightarrow \boxed{\text{B}} <br />
\end{align*}</cmath><br />
<br />
===Solution 3===<br />
<br />
We can use the formula for finite arithmetic sequences.<br />
<br />
It is <math>\frac{n}{2}\times</math> (First Term+Last Term) where <math>n</math> is the number of terms in the sequence. <br />
<br />
Applying it here:<br />
<br />
<math>\frac{10}{2} \times (90+99) = 945 \rightarrow \boxed{B}</math><br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1985|num-b=1|num-a=3}}<br />
<br />
<br />
{{MAA Notice}}<br />
[[Category:Introductory Algebra Problems]]</div>Peashooter6969https://artofproblemsolving.com/wiki/index.php?title=1985_AJHSME_Problems/Problem_2&diff=943261985 AJHSME Problems/Problem 22018-05-07T00:46:17Z<p>Peashooter6969: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
<br />
<math>90+91+92+93+94+95+96+97+98+99=</math><br />
<br />
<br />
<math>\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045</math><br />
<br />
==Solution==<br />
<br />
===Solution 1===<br />
One possibility is to simply add them. However, this can be time-consuming, and there are other ways to solve this problem. <br />
We find a simpler problem in this problem, and simplify -> <math>90 + 91 + ... + 98 + 99 = 90 \times 10 + 1 + 2 + 3 + ... + 8 + 9</math><br />
<br />
We know <math>90 \times 10</math>, that's easy - <math>900</math>. So how do we find <math>1 + 2 + ... + 8 + 9</math>?<br />
<br />
We rearrange the numbers to make <math>(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5</math>. You might have noticed that each of the terms we put next to each other add up to 10, which makes for easy adding. <math>4 \times 10 + 5 = 45</math>. Adding that on to 900 makes 945.<br />
<br />
945 is <math>\boxed{\text{B}}.</math><br />
<br />
===Solution 2===<br />
Instead of breaking the sum and then rearranging, we can start by rearranging:<br />
<cmath>\begin{align*}<br />
90+91+92+\cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\<br />
&= 189+189+189+189+189 \\<br />
&= 945\rightarrow \boxed{\text{B}} <br />
\end{align*}</cmath><br />
<br />
===Solution 3===<br />
<br />
We can use a formula. <br />
<br />
It is <math>\frac{n}{2}\times</math> (First term+Last term) where <math>n</math> is the number of terms in the sequence. <br />
<br />
Applying it here:<br />
<br />
<math>\frac{10}{2} \times (90+99) = 945 \rightarrow \boxed{B}</math><br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1985|num-b=1|num-a=3}}<br />
<br />
<br />
{{MAA Notice}}<br />
[[Category:Introductory Algebra Problems]]</div>Peashooter6969https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_4&diff=943252018 AMC 10A Problems/Problem 42018-05-07T00:39:22Z<p>Peashooter6969: /* See Also */</p>
<hr />
<div>==Problem==<br />
<br />
How many ways can a student schedule 3 mathematics courses -- algebra, geometry, and number theory -- in a 6-period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other 3 periods is of no concern here.)<br />
<br />
<math>\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24</math><br />
<br />
==Solution 1==<br />
<br />
We must place the classes into the periods such that no two classes are in the same period or in consecutive periods.<br />
<br />
Ignoring distinguishability, we can thus list out the ways that three periods can be chosen for the classes when periods cannot be consecutive:<br />
<br />
Periods <math>1, 3, 5</math><br />
<br />
Periods <math>1, 3, 6</math><br />
<br />
Periods <math>1, 4, 6</math><br />
<br />
Periods <math>2, 4, 6</math><br />
<br />
There are <math>4</math> ways to place <math>3</math> nondistinguishable classes into <math>6</math> periods such that no two classes are in consecutive periods. For each of these ways, there are <math>3! = 6</math> orderings of the classes among themselves.<br />
<br />
Therefore, there are <math>4 \cdot 6 = \boxed{\textbf{(E) } 24}</math> ways to choose the classes.<br />
<br />
-Versailles15625<br />
<br />
==Solution 2==<br />
Realize that the number of ways of placing, regardless of order, the 3 mathematics courses in a 6-period day so that no two are consecutive is the same as the number of ways of placing 3 mathematics courses in a sequence of 4 periods regardless of order and whether or not they are consecutive.<br />
<br />
To see that there is a one to one correlation, note that for every way of placing 3 mathematics courses in 4 total periods (as above) one can add a non-mathematics course between each pair (2 total) of consecutively occurring mathematics courses (not necessarily back to back) to ensure there will be no two consecutive mathematics courses in the resulting 6-period day.<br />
For example, where <math>M</math> denotes a math course and <math>O</math> denotes a non-math course:<br />
<math>M O M M \rightarrow M O O M O M</math><br />
<br />
For each 6-period sequence consisting of <math>M</math>s and <math>O</math>s, we have <math>3!</math> orderings of the 3 distinct mathematics courses.<br />
<br />
So, our answer is <math>\dbinom{4}{3}(3!)= \boxed{\textbf{(E) } 24}</math><br />
<br />
- Gregwwl<br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=A|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Peashooter6969