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2017 AMC 8 Problems/Problem 24
==Solution 1 (Principle of Inclusion-Exclusion)== We use Principle of Inclusion-Exclusion. There are <math>365</math> days in the year, and we subtract the

6 KB (907 words) - 21:11, 20 January 2024
The Pot Method
The Pot Method is a double-counting method used to evaluate series with binomial coefficients, especial The other way to count this expected value is case-by-case based on the number of blue marbles.

3 KB (414 words) - 19:10, 4 April 2019
MIE 2015
...ree sets <math>F</math>, <math>G</math> and <math>H</math>. The set <math>G-H</math> is equal to the set: (a) <math>(G\cup F)-(F-H)</math>

7 KB (1,127 words) - 18:23, 11 January 2018
2018 AMC 10A Problems/Problem 20
...y it in particular. Each of these sections has <math>10</math> distinct sub-squares, whether partially or in full. So since each can be colored either w .... Since we must subtract off <math>2</math> cases for the all-black and all-white cases, the answer is <math>2^{10}-2=\boxed{\textbf{(B) } 1022.}</math>

5 KB (827 words) - 21:31, 24 October 2023
2018 AMC 10A Problems/Problem 22
== Video Solution by OmegaLearn (Meta-Solving Technique) == {{AMC10 box|year=2018|ab=A|num-b=21|num-a=23}}

10 KB (1,599 words) - 04:51, 6 August 2023
2018 AMC 10A Problems/Problem 21
...math>a</math> for which the curves <math>x^2+y^2=a^2</math> and <math>y=x^2-a</math> in the real <math>xy</math>-plane intersect at exactly <math>3</mat Substituting <math>y=x^2-a</math> into <math>x^2+y^2=a^2</math>, we get

9 KB (1,502 words) - 23:31, 19 August 2023
2018 AMC 12A Problems/Problem 19
{{AMC12 box|year=2018|ab=A|num-b=18|num-a=20}}

7 KB (1,147 words) - 20:27, 3 July 2024
2018 AMC 12B Problems
...sqrt2 \qquad \textbf{(D) } \frac{1}{2}\sqrt{2} \qquad \textbf{(E) } \sqrt 3-1</math> ...ac{4}{7}</cmath>and <math>q</math> is as small as possible. What is <math>q-p</math>?

14 KB (2,118 words) - 15:36, 28 October 2021
2018 AMC 10B Problems/Problem 19
...>p^2q^2.</math> Since <math>c-1<100,</math> the only possibility is <math>c-1=2^2\cdot3^2=36,</math> from which <math>c=37.</math> We conclude that Joey ...</math>. Now, since <math>j-1=37</math>, by similar logic, <math>37=(1+k)(a-1)</math>, so <math>k=36</math> and Joey will be <math>38+36=74</math> and t

5 KB (888 words) - 05:20, 12 April 2024
2018 AMC 10B Problems/Problem 16
...ath> consecutive numbers must be a multiple of <math>k</math>, so <math>a^3-a</math> is both divisible by <math>2</math> and <math>3</math>. This provid {{AMC10 box|year=2018|ab=B|num-b=15|num-a=17}}

5 KB (814 words) - 19:27, 5 March 2024
2018 AMC 10B Problems
A three-dimensional rectangular box with dimensions <math>X</math>, <math>Y</math>, for(int j=0;j<=3-i;j=j+1)

15 KB (2,237 words) - 23:08, 15 November 2023
2018 AMC 12B Problems/Problem 9
<ol style="margin-left: 1.5em;"> More generally, the term <math>k+101</math> occurs <math>100-k</math> times for <math>k\in\{1,2,3,\ldots,99\}.</math><p></li>

5 KB (594 words) - 22:03, 27 May 2023
2018 AMC 10B Problems/Problem 8
for(int j=0;j<=3-i;j=j+1) This is a 3-step staircase and uses 18 toothpicks. How many steps would be in a staircas

4 KB (609 words) - 19:49, 27 August 2023
2018 USAJMO Problems/Problem 5
...th>\{1, 2, ..., p\},</math> so they have a total of exactly <math>\frac{p(p-1)}{2}</math> edges. Therefore, there exists at least one graph <math>G_k</m {{USAJMO newbox|year=2018|num-b=4|num-a=6}}

2 KB (455 words) - 13:53, 21 April 2018
2018 USAMO Problems/Problem 4
...th>\{1, 2, ..., p\},</math> so they have a total of exactly <math>\frac{p(p-1)}{2}</math> edges. Therefore, there exists at least one graph <math>G_k</m The first equality is given by a well known theorem, which can be proven by C-S or Jensen's.

4 KB (794 words) - 21:46, 20 March 2023
2018 USAMO Problems/Problem 3
s+1. Therefore, the desired sum of k-th powers is equal is a complete reduced residue set modulo nq. The new sum of k-th powers is equal to

5 KB (939 words) - 22:00, 11 February 2024
2018 USAMO Problems/Problem 6
Now, consider what happens when the two rows are swapped (and the top-bottom pairs are reordered so that the top reads (1,2,3,...,n)). This will ...ds <math>b_n=n\cdot b_{n-1}</math>* which has the same parity as <math>b_{n-1}</math>, so we need only consider the parity of <math>b_n</math> for odd <

4 KB (700 words) - 04:36, 5 March 2023
Lagrange's Identity
...\biggr )}-{\biggl (}\sum _{k=1}^{n}a_{k}b_{k}{\biggr )}^{2}&=\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}\\&{\biggl (}={\frac {1}{2}}\s The vector form follows from the Binet-Cauchy identity by setting <math>ci = ai</math> and <math>di = bi</math>. Th

4 KB (914 words) - 17:27, 31 August 2022
1960 AHSME Problems/Problem 34
...answer. From the graph (where the x-axis is the time in seconds and the y-axis is distance from one side of the pool), there are five meeting points, {{AHSME 40p box|year=1960|num-b=33|num-a=35}}

2 KB (262 words) - 19:18, 17 May 2018
2018 AIME I Problems
Find the number of four-element subsets of <math>\{1,2,3,4,\dots, 20\}</math> with the property that ...at when <math>3^n</math> is written in base <math>143</math>, its two right-most digits in base <math>143</math> are <math>01</math>.

8 KB (1,284 words) - 14:35, 9 August 2021

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