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remainer

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remainer
▪ I. † reˈmainer1 Obs. Also 5 remaner, 5–6 remayner, 7 remainor. [a. ONF. remaneir, = OF. remanoir inf.: see remain v. and -er4.] 1. Law. a. ? = remanet 2 a. rare—1.1454 Paston Lett. I. 294 Mastere Pownyngs hath day tille the next terme by a remayner. b. = remainder n.1 1.1473 Rolls of Parlt. VI. 75... Oxford English Dictionary
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Remain
Remain" (a member of the European Union), one of the two options available to voters in the 2016 United Kingdom European Union membership referendum Remainer wikipedia.org
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remaner
remaner obs. form of remainer. Oxford English Dictionary
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Thomas Borwick
Douglas Carswell, former UKIP MP; and Voter Consultancy Ltd, whose micro-targeting of Facebook Ads allegedly precipitated unforeseen death threats against Remainer wikipedia.org
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What is the number that, when divided by $3$, $5$, $7$, leaves remainders of $2$, $3$, $2$, respectively? What is the number? The LCM of the divisors is 105. I think this has something to do with the Chinese remainer...
Yes, this is the first known incidence of the **Chinese Remainder Theorem** (CRT). From Wikipedia: > The earliest known statement of the CRT, as a problem with specific numbers, appears in the 3rd-century book Sunzi's Mathematical Classic () by the Chinese mathematician Sun Tzu:[2] “ There are certa...
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Wardell Westby
Westby's widow died in Great Marlborough Street on 10 February 1760, and of his daughter it was said ‘Let the remainer of her unhappy story be left in wikipedia.org
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remayne
† reˈmayne, v. Obs. rare. [ad. OF. remainer, -me(i)ner, etc., f. re- re + mener to lead: cf. mod.F. ramener.] trans. To lead or bring back.1481 Caxton Myrr. i. xii. 37 Musyque accordeth alle thinges that dyscorde..& remayne[th] them to concordaunce. Ibid. 38. Oxford English Dictionary
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Laura Quevedo
Back in Spain, she played with Pajariel Bembibre for the remainer of the 2014-2015 season. wikipedia.org
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substander
† subˈstander Obs. [Rendering of L. substans (see substance n.).] A thing that subsists. So subˈstanding ppl. a., subsisting.1662 J. Chandler Van Helmont's Oriat. 144 A truly substanding or remaining Being [orig. vere substantis entis]. Ibid. 345 The Substance of that Substander or remainer [orig. e... Oxford English Dictionary
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minorand
† minorand Arith. Obs. [ad. med.L. minōrandus (sc. numerus), gerundive of minōrāre: see next.] = minuend.1709–29 V. Mandey Syst. Math., Arith. 13 The Remainer added to the Subducend, if the Sum makes the Minorand, 'tis right. Oxford English Dictionary
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Binary division, with reminder I'm trying to do a binary division to find a remainder. Here's what I've done: 10011100/1001 = 10001 1001 ---- 00001 0 ----- 11 0 ...
Both calculators are computing in GF(2), the Galois field of order $2$. In this field, subtraction does not involve carries -- subtraction is accomplished by XORing the operands. For instance, at your first link, the last "subtraction" is "$1100 \underline{\vee} 1001 = 101$. An actual binary calcula...
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Conclude there are infinitely primes with given residue class Things I know (already proved): 1. There are infinitely many primes $p$, such that $p \equiv 1 \pmod{3}$ 2. There are infinitely many primes $p$, such...
Yes, it is sort of possible except that it requires one additional piece of theory (not as strong as Dirichlet's theorem), namely quadratic reciprocity. Given any finite set of primes $p_1,p_2,\ldots, p_k$, each of the form $5k-1$, we can form the value $$N = 5(2p_1p_2\cdots p_k)^2 - 1.$$ It is clea...
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If $x$ is a square modulo two primes, then it is a square modulo their product > $a, b$ be integers, $p, q$ primes. If $x \equiv a^2 $ (mod $p$) and $x \equiv b^2$ (mod $q$), then $x \equiv c^2$ (mod $pq$) for some in...
By the Chinese Remainder Theorem, there exists an element $c$ such that $$c \equiv a \pmod{p} \\\ c \equiv b \pmod{q}$$ Then $$x \equiv c^2 \pmod{p}\\\ x\equiv c^2 \pmod{q}$$
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Solve congruence with large exponents I'm trying to solve: > $$ x^{1477} \equiv 54 \mod 97 $$ Applying Euler-Fermat gives: $$ x^{1477} = x^{15\cdot 96 + 37} = x^{37}\cdot x^{{96}^{15}} \equiv x^{37}\cdot 1^{15} = x...
If $x^{37} \equiv 54 \pmod {97}$ then $x^{37\cdot 24} \equiv 54^{24} \equiv 1 \pmod {97}$ because the order of $54 \pmod {97}$ is $24$. This means the order of $x$ is a common divisor of $96$ and $37 \cdot 24$. Therefore the order of $x$ is a divisor of $24$, and so $x^{24} \equiv 1 \pmod {97}$. But...
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Why does m dividing a − b mean that a and b have the same remainer under division by m? Why does $m$ dividing $a − b$ mean that $a$ and $b$ have the same remainer under division by $m$? By the definition of a “remain...
Continuing from where you left off, this means that there is some $t$ such that: $m(i-j) + r_1 - r_2 = mt$, so $r_1 - r_2 = m(t-i+j)$, hence $m$ divides $r_1-r_2$. But $0 \le r_1 < m$ and $0 \le r_2 < m$, so $|r_1 - r_2| < m$. This can only mean that $r_1 - r_2 = 0$.
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