Search results

Create the page "1000" on this wiki! See also the search results found.

1991 AIME Problems
Expanding <math>(1+0.2)^{1000}_{}</math> by the binomial theorem and doing no further manipulation gives ...\choose 1}(0.2)^1+{1000 \choose 2}(0.2)^2+\cdots+{1000 \choose 1000}(0.2)^{1000}</math></center>

7 KB (1,106 words) - 22:05, 7 June 2021
1992 AIME Problems
For how many pairs of consecutive integers in <math>\{1000,1001,1002^{}_{},\ldots,2000\}</math> is no carrying required when the two i

8 KB (1,117 words) - 05:32, 11 November 2023
1993 AIME Problems
...rawn randomly and without replacement from the set <math>\{1, 2, 3,\ldots, 1000\}</math>. Three other numbers, <math>b_1, b_2, b_3</math>, are then drawn r

8 KB (1,275 words) - 06:55, 2 September 2021
1994 AIME Problems
...What is the remainder when the 1994th term of the sequence is divided by 1000? ...94,\,</math> what is the remainder when <math>f(94)\,</math> is divided by 1000?

7 KB (1,141 words) - 07:37, 7 September 2018
1996 AIME Problems
...th>. For how many positive integers <math>n</math> is it true that <math>n<1000</math> and that <math>\lfloor \log_{2} n \rfloor</math> is a positive even

6 KB (931 words) - 17:49, 21 December 2018
1997 AIME Problems
How many of the integers between 1 and 1000, inclusive, can be expressed as the difference of the squares of two nonneg

7 KB (1,098 words) - 17:08, 25 June 2020
1998 AIME Problems
Except for the first two terms, each term of the sequence <math>1000, x, 1000 - x,\ldots</math> is obtained by subtracting the preceding term from the on

7 KB (1,084 words) - 02:01, 28 November 2023
1999 AIME Problems
...whose labels divide the label on the <math>i</math>-th switch. After step 1000 has been completed, how many switches will be in position <math>A</math>?

7 KB (1,094 words) - 13:39, 16 August 2020
2000 AIME I Problems
...</math> and <math>b</math> are relatively prime positive divisors of <math>1000.</math> What is the greatest integer that does not exceed <math>\frac{S}{10

7 KB (1,204 words) - 03:40, 4 January 2023
2003 AIME I Problems
An integer between <math>1000</math> and <math>9999,</math> inclusive, is called balanced if the sum of i ...> m </math> and <math> n </math> are positive integers with <math> m + n < 1000, </math> find <math> m + n. </math>

6 KB (965 words) - 16:36, 8 September 2019
2002 AIME II Problems
...sitive integer. Find the remainder when <math>m</math> is divided by <math>1000</math>. Find the integer that is closest to <math>1000\sum_{n=3}^{10000}\frac1{n^2-4}</math>.

7 KB (1,177 words) - 15:42, 11 August 2023
2003 AIME II Problems
...are all different. What is the remainder when <math>N</math> is divided by 1000?

7 KB (1,127 words) - 09:02, 11 July 2023
1983 AIME Problems/Problem 10
...0</math> to <math>1999</math>), so by complementary counting you get <math>1000-(504+27+36+1)=\boxed{432}</math> numbers.

5 KB (855 words) - 20:26, 14 January 2023
1984 AIME Problems/Problem 7
n-3&\mbox{if}\ n\ge 1000\\ f(f(n+5))&\mbox{if}\ n<1000\end{cases}</math>

4 KB (617 words) - 22:09, 15 May 2024
1984 AIME Problems/Problem 12
...roots <math>f(x)=0</math> must have in the interval <math>-1000\leq x \leq 1000</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page In the interval <math>-1000\leq x\leq 1000</math>, there are <math>201</math> multiples of <math>10</math> and <math>2

3 KB (588 words) - 14:37, 22 July 2020
1985 AIME Problems/Problem 10
How many of the first 1000 [[positive integer]]s can be expressed in the form ...rs and so we hit <math>50 \cdot 12 = \boxed{600}</math> of the first <math>1000</math>.

12 KB (1,859 words) - 18:16, 28 March 2022
1985 AIME Problems/Problem 7
...- t^2 = 1</math>. Then <math>s = 10, t = 3</math> and so <math>d = s^3 = 1000</math>, <math>b = t^5 = 243</math> and <math>d-b=\boxed{757}</math>.

1 KB (222 words) - 11:04, 4 November 2022
1986 AIME Problems/Problem 10
Since <math>0<100a+10b+c<1000</math>, we get the inequality <cmath>N<222(a+b+c)<N+1000</cmath>

3 KB (565 words) - 16:51, 1 October 2023
1986 AIME Problems/Problem 5
...such that <math>n+10 \mid n^3 +100</math>, we have: <math>n+10 \mid ((n^3 +1000) - (n^3 +100) \longrightarrow n+10 \mid 900</math>. This is because of the

2 KB (338 words) - 19:56, 15 October 2023
1987 AIME Problems/Problem 12
...r]] and <math>r</math> is a [[positive]] [[real number]] less than <math>1/1000</math>. Find <math>n</math>. ...math>, so it is possible for <math>r</math> to be less than <math>\frac{1}{1000}</math>. However, we still have to make sure a sufficiently small <math>r<

4 KB (673 words) - 19:48, 28 December 2023
1987 AIME Problems/Problem 7
...riples]] <math>(a,b,c)</math> of positive integers for which <math>[a,b] = 1000</math>, <math>[b,c] = 2000</math>, and <math>[c,a] = 2000</math>. <math>1000 = 2^35^3</math> and <math>2000 = 2^45^3</math>. By [[LCM#Using prime factor

3 KB (547 words) - 22:54, 4 April 2016
1988 AIME Problems/Problem 9
...of <math>(100k + 42)^3 \equiv 3(100k)(42)^2 + 42^3 \equiv 200k + 88 \pmod{1000}</math>. Hence the lowest possible value for the hundreds digit is <math>4< ...is <math>(100k + 92)^3 \equiv 3(100k)(92)^2 + 92^3 \equiv 200k + 688 \pmod{1000}</math>. The lowest possible value for the hundreds digit is <math>1</math>

6 KB (893 words) - 08:15, 2 February 2023
1989 AIME Problems/Problem 11
A sample of 121 [[integer]]s is given, each between 1 and 1000 inclusive, with repetitions allowed. The sample has a unique [[mode]] (most ...ish to minimize or maximize <math>x</math> (in other words, <math>x \in [1,1000]</math>). Indeed, <math>D(x)</math> is symmetric about <math>x = 500.5</mat

5 KB (851 words) - 18:01, 28 December 2022
1989 AIME Problems/Problem 9
&\equiv893\pmod{1000}. &\equiv224\pmod{1000}.

6 KB (874 words) - 15:50, 20 January 2024
1989 AIME Problems/Problem 3
...ometric series]], <math>0.d25d25d25\ldots = \sum_{n = 1}^\infty \frac{d25}{1000^n} = \frac{100d + 25}{999}</math>. Thus <math>\frac{n}{810} = \frac{100d + To get rid of repeating decimals, we multiply the equation by 1000. We get <math>\frac{1000n}{810} = d25.d25d25...</math> We subtract the orig

3 KB (499 words) - 22:17, 29 March 2024
1990 AIME Problems/Problem 3
...y conceivable reasoning behind this is that <math>r</math> is greater than 1000. This prompts us to look into the second case, where <math>s</math> divides

3 KB (516 words) - 19:18, 16 April 2024
1991 AIME Problems/Problem 3
Expanding <math>(1+0.2)^{1000}_{}</math> by the binomial theorem and doing no further manipulation gives ...\choose 1}(0.2)^1+{1000 \choose 2}(0.2)^2+\cdots+{1000 \choose 1000}(0.2)^{1000}</math>

5 KB (865 words) - 12:13, 21 May 2020
1992 AIME Problems/Problem 6
For how many pairs of consecutive integers in <math>\{1000,1001,1002,\ldots,2000\}</math> is no carrying required when the two integer

3 KB (455 words) - 02:03, 10 July 2021
1992 AIME Problems/Problem 3
...ath>\frac{n}{2n}=\frac{1}{2}</math>, and <math>\frac{n+3}{2n+4}>\frac{503}{1000}</math>. ...3}{2n+4} > .503 = \frac{503}{1000}.</cmath>Cross-multiplying, we get <math>1000(n+3) > 503(2n+4),</math> which is equivalent to <math>n < \frac{988}{6} = 1

2 KB (251 words) - 08:05, 2 January 2024
1993 AIME Problems/Problem 11
...math>a_6=\frac{364}{729}</math>, <math>m+n = 1093 \equiv \boxed{093} \pmod{1000}</math>.

7 KB (1,058 words) - 20:57, 22 December 2020
1993 AIME Problems/Problem 7
...rawn randomly and without replacement from the set <math>\{1, 2, 3,\ldots, 1000\}</math>. Three other numbers, <math>b_1, b_2, b_3</math>, are then drawn r There is a total of <math>P(1000,6)</math> possible ordered <math>6</math>-tuples <math>(a_1,a_2,a_3,b_1,b_2

5 KB (772 words) - 09:04, 7 January 2022
1994 AIME Problems/Problem 3
.../math> what is the remainder when <math>f(94)\,</math> is divided by <math>1000</math>?

2 KB (252 words) - 11:12, 3 July 2023
1994 AIME Problems/Problem 1
...at is the [[remainder]] when the 1994th term of the sequence is divided by 1000? ...97-1) \cdot 3 = 2992</math>. The value of <math>n^2 - 1 = 2992^2 - 1 \pmod{1000}</math> is <math>\boxed{063}</math>.

946 bytes (139 words) - 21:05, 1 September 2023
1995 AIME Problems/Problem 2
...95^2</math>, making the solution <math>(2000-5)^2 \equiv \boxed{025} \pmod{1000}</math>. ...^2\pmod{1000}\equiv 995^2\pmod{1000}\equiv (-5)^2\pmod{1000}\equiv 25\pmod{1000}</math>, so our answer is <math>\boxed{025}</math>.

2 KB (362 words) - 00:40, 29 January 2021
1996 AIME Problems/Problem 15
...5^{\circ}</math>. The answer is <math>\lfloor 1000r \rfloor = \left\lfloor 1000 \cdot \frac{180 - 5\theta}{180 - 3\theta} \right\rfloor = \left \lfloor \fr

5 KB (710 words) - 21:04, 14 September 2020
1996 AIME Problems/Problem 2
...d x. For how many positive integers <math>n</math> is it true that <math>n<1000</math> and that <math>\lfloor \log_{2} n \rfloor</math> is a positive even ...h>n</math> must satisfy these [[inequality|inequalities]] (since <math>n < 1000</math>):

1 KB (163 words) - 19:31, 4 July 2013
1997 AIME Problems/Problem 3
...y-\dfrac{1000}{9}\right)=\dfrac{1000}{9}</math>, and <math>(9x-1)(9y-1000)=1000</math>. Since <math>89 < 9x-1 < 890</math>, we can use trial and error on factors of 1000. If <math>9x - 1 = 100</math>, we get a non-integer. If <math>9x - 1 = 125<

2 KB (375 words) - 19:34, 4 August 2021
1997 AIME Problems/Problem 1
How many of the integers between 1 and 1000, inclusive, can be expressed as the [[difference of squares|difference of t

801 bytes (115 words) - 15:52, 2 March 2020
1998 AIME Problems/Problem 8
Except for the first two terms, each term of the sequence <math>1000, x, 1000 - x,\ldots</math> is obtained by subtracting the preceding term from the on ...math><font color="white">aaa</font> || <math>1000 - x</math> || <math>2x - 1000</math><font color="white">a</font> || <math>2000 - 3x</math> || <math>5x -

2 KB (354 words) - 19:37, 24 September 2023
1999 AIME Problems/Problem 7
...whose labels divide the label on the <math>i</math>-th switch. After step 1000 has been completed, how many switches will be in position <math>A</math>? The number of switches in position A is <math>1000-125-225 = \boxed{650}</math>.

3 KB (475 words) - 13:33, 4 July 2016
2000 AIME I Problems/Problem 15
...aining, some cards have not even made one trip through yet, <math>2(1024 - 1000) = 48</math>, to be exact. Once these cards go through, 1999 will be the <m ...s initially in the deck once, in round two, you would go through all <math>1000</math> cards initially in the deck once, so on and so forth. For each round

15 KB (2,673 words) - 19:16, 6 January 2024
2000 AIME I Problems/Problem 14
...<math>\angle ACB = 80^{\circ}</math>, so the answer is <math>\left\lfloor 1000 \cdot \frac{80}{140} \right\rfloor = \left\lfloor \frac{4000}{7} \right\rfl <math>\left\lfloor 1000\left(\frac {4}{7}\right)\right\rfloor = \boxed{571}</math>.

8 KB (1,275 words) - 03:04, 27 February 2022
2000 AIME I Problems/Problem 11
...and <math>b</math> are [[relatively prime]] positive [[divisor]]s of <math>1000.</math> What is the [[floor function|greatest integer]] that does not excee Since all divisors of <math>1000 = 2^35^3</math> can be written in the form of <math>2^{m}5^{n}</math>, it f

4 KB (667 words) - 13:58, 31 July 2020
2000 AIME I Problems/Problem 9
...ifferent author: <math>-(3 - \log 2000) = \log 2000 - 3 = \log 2000 - \log 1000 = \log 2.</math>

4 KB (623 words) - 15:56, 8 May 2021
2000 AIME I Problems/Problem 6
...\sqrt{10^6} = 10^3</math>, then we can use all positive integers less than 1000 for <math>a</math> and <math>b</math>. ...square, <math>y</math> must also be a perfect square. Since <math>0 < y < (1000)^2</math>, <math>y</math> must be from <math>1^2</math> to <math>999^2</mat

6 KB (966 words) - 21:48, 29 January 2024
2001 AIME I Problems/Problem 8
Given this is an AIME problem, <math>A<1000</math>. If we look at <math>B</math> in base <math>10</math>, it must be eq

3 KB (502 words) - 11:28, 9 December 2023
2001 AIME I Problems/Problem 3
...x</math> is a root, <math>x</math> is also a root. Thus, we pair up <math>1000</math> pairs of roots that sum to <math>\frac{1}{2}</math> to get a sum of

2 KB (335 words) - 18:38, 9 February 2023
2002 AIME I Problems/Problem 7
An equivalent statement is to note that we are looking for <math>1000 \left\{\frac{10^{859}}{7}\right\}</math>, where <math>\{x\} = x - \lfloor x

2 KB (316 words) - 19:54, 4 July 2013
2003 AIME I Problems/Problem 14
<center><math>\frac{251}{1000} \le \frac{m'}{n} < \frac{252}{1000} \Longleftrightarrow 251n \le 1000m' < 252n \Longleftrightarrow n \le 250(4 ...ger <math>m</math> such that <math>0<\frac{m}{n}-\frac{251}{1000}<\frac{1}{1000}</math>.

3 KB (477 words) - 14:23, 4 January 2024
2003 AIME I Problems/Problem 13
.../math>'s. Find the [[remainder]] when <math> N </math> is divided by <math>1000</math>.

4 KB (651 words) - 19:42, 7 October 2023

Page

Toolbox

Search

Search results