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- ...extrm{ } (\textrm{mod }n)</math>, or <cmath>a \equiv b \textrm{ } (\textrm{mod }n).</cmath> Therefore, we wish to show that there exist <math>a</math> and ...math>\frac{p}{q}</math> with <math>1 \leq q \leq n</math> such that <math>|x - \frac{p}{q}| < \frac{1}{nq}</math>.''11 KB (1,985 words) - 21:03, 5 August 2023
- Taken <math>\mod p</math>, all of the middle terms disappear, and we end up with <math>(a+1) Thus, <math>\mod{p}</math>, we have that the product of the elements of <math>S</math> is16 KB (2,675 words) - 10:57, 7 March 2024
- ...math>. <math>3</math> is a divisor of the [[LHS]] (also notice that <math>x-3y</math> must always be an integer). However, <math>17</math> will never An equation of form <math>x^4+y^4=z^2</math> has no [[integer]] solutions, as follows:9 KB (1,434 words) - 13:10, 20 February 2024
- The '''(mod 5)''' part just tells us that we are working with the integers modulo 5. I ...math>7^{7^{1000}}</math>. After that, we see that 7 is congruent to -1 in mod 4, so we can use this fact to replace the 7s with -1s, because 7 has a patt15 KB (2,396 words) - 20:24, 21 February 2024
- ...to'' <math>b</math> ''modulo'' <math>n</math>, or <math>a \equiv b</math> (mod <math>n</math>), if the difference <math>{a - b}</math> is divisible by <ma ...d '''the integers modulo <math>n</math>''' (usually known as "the integers mod <math>n</math>," or <math>\mathbb{Z}_n</math> for short). This structure g14 KB (2,317 words) - 19:01, 29 October 2021
- <cmath>(x+y+z)^{2006}+(x-y-z)^{2006}</cmath> <cmath>\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^ay^bz^c}</cmath>8 KB (1,332 words) - 17:37, 17 September 2023
- ...^2</math>, other than <math>1</math>, for integers <math>x,y</math> (<math>x</math> and <math>y</math> can be equal), ie. <math>4,8,9,12,16,18,20,24,25, ...the ratio be 7 and the common difference to be 6. We see multiplying by 7 mod 12 that the geometric sequence is alternating from 1 to 7 to 1 to 7 and so4 KB (792 words) - 00:29, 13 April 2024
- ...nd only if either <math>x + y = \frac \pi 2 + 2\pi \cdot k</math> or <math>x - y = \frac\pi2 + 2\pi\cdot k</math> for some integer <math>k</math>. So f ...get <math>90n \equiv 90 \mod 360</math>. Canceling gets <math>n \equiv 1 \mod 4</math>, and thus there are <math>\boxed{250}</math> values of n.6 KB (1,154 words) - 03:30, 11 January 2024
- ...r example, with the binary string 0001001000 <math>y</math> is 6 and <math>x</math> is 3 (note that it is zero indexed). ...h>x</math> and <math>y</math> are <math>0 \leq x < y < 40</math> and <math>x, y \in \mathbb{Z}</math>.7 KB (1,091 words) - 18:41, 4 January 2024
- ...ber of 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 If <math>f(2+x)=f(2-x)</math>, then substituting <math>t=2+x</math> gives <math>f(t)=f(4-t)</math>. Similarly, <math>f(t)=f(14-t)</math>3 KB (588 words) - 14:37, 22 July 2020
- Take an even positive integer <math>x</math>. <math>x</math> is either <math>0 \bmod{6}</math>, <math>2 \bmod{6}</math>, or <math If <math>x \ge 18</math> and is <math>0 \bmod{6}</math>, <math>x</math> can be expressed as <math>9 + (9+6n)</math> for some nonnegative <ma8 KB (1,346 words) - 01:16, 9 January 2024
- ...>(x+x^2+x^3)^7</math>. Our goal is to find the coefficients of every <math>x^{4n}</math> and add them up before dividing by <math>3^7</math>. Here we ha Let <math>\omega = e^{i\pi / 2}</math>. We have that if <math>G(x) = (x+x^2+x^3)^7</math>, then <cmath>\frac{G(1) + G(\omega) + G(\omega^2) + G(\omega^3)17 KB (2,837 words) - 13:34, 4 April 2024
- ...600k^2 + 120k + 8</math>. Since we are looking for the tens digit, <math>\mod{100}</math> we get <math>20k + 8 \equiv 88 \pmod{100}</math>. This is true ...positive integer. Then, this becomes <math>y^3 = 125a + 111</math>. Taking mod <math>5</math>, <math>25</math>, and <math>125</math>, we get <math>y^3 \eq6 KB (893 words) - 08:15, 2 February 2023
- ...form of <math>5^2 \cdot y^3</math> based upon the second part, with <math>x</math> and <math>y</math> denoting an [[integer]]s. <math>c</math> is minim <math>k \equiv 0 \text{(mod 2)}</math>3 KB (552 words) - 12:41, 3 March 2024
- ...f set <math>B</math> as <math>\{n^3, n^6, \ldots n^{144}\}</math>. <math>n^x</math> can yield at most <math>144</math> different values. All solutions f ...,8</math>. Since the exponents are of roots of unities, they reduce <math>\mod{144}</math>, so all numbers in the range are covered. Thus the answer is <m3 KB (564 words) - 04:47, 4 August 2023
- ...ign*}9x &= a \\ 11y &= a \\ 10z + 5 &= a, \end{align*}</cmath> here, <math>x,</math> and <math>y</math> are the middle terms of the sequence of 9 and 11 ...<math>x,x+1, \cdots x+9.</math> We have <math>10x+45 = 99k.</math> Taking mod <math>5</math> yields <math>k \equiv 0 \pmod{5}.</math> Since <math>k</math3 KB (524 words) - 18:06, 9 December 2023
- ...th <math>y<x\le 100,</math> are both <math>\frac xy</math> and <math>\frac{x+1}{y+1}</math> integers? ...lso, as <math>\text{gcd}\,(y,y+1) = 1</math>, it follows that <math>y(y+1)|x-y</math>. {{ref|1}}4 KB (646 words) - 17:37, 1 January 2024
- .../math> the arithmetic mean of the set of values obtained by deleting <math>x</math> from <math>\mathcal{S}</math> is an integer. Given that 1 belongs t ...ath>n^2+1</math>, the <math>(n+1)</math>th positive integer congruent to 1 mod <math>n</math>.2 KB (267 words) - 19:18, 21 June 2021
- ...d <math>n</math> is small as possible. Then <math>10^k \cdot \frac{m}{n} - X = \frac{10^k m - nX}{n} = 0.251 \ldots</math>. Since <math>10^k m - nX</mat ...> mod <math>1000</math>. We see that <math>n</math> must be <math>3</math> mod <math>4</math> to have this happen (as this reduces the distance between th3 KB (477 words) - 14:23, 4 January 2024
- ...is implies <math>3|d</math> since all other terms are congruent to <math>0\mod 3</math>. Let <math>9a^2+120a=x^2</math>, where <math>x</math> is an integer. This yields the following:5 KB (921 words) - 23:21, 22 January 2023
