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Number TheoryQuestion and Answers: Page 9

Question Number 126269 Answers: 1 Comments: 0

If n is an odd integer , show that 32 divides (n^2 +3)(n^3 +7) ?

$${If}\:{n}\:{is}\:{an}\:{odd}\:{integer}\:,\:{show}\:{that}\:\mathrm{32} \\ $$$${divides}\:\left({n}^{\mathrm{2}} +\mathrm{3}\right)\left({n}^{\mathrm{3}} +\mathrm{7}\right)\:?\: \\ $$

Question Number 125959 Answers: 1 Comments: 0

Question Number 125666 Answers: 0 Comments: 0

solution : x′ = (((−2 1)),((1 −2)) )x + (((2e^(−m) )),((3m)) )

$${solution}\:: \\ $$$$ \\ $$$${x}'\:=\:\begin{pmatrix}{−\mathrm{2}\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\:\:−\mathrm{2}}\end{pmatrix}{x}\:+\:\begin{pmatrix}{\mathrm{2}{e}^{−{m}} }\\{\mathrm{3}{m}}\end{pmatrix} \\ $$

Question Number 125042 Answers: 0 Comments: 3

2+2+2=6 3×3−3=6 (√4)+(√4)+(√4)=6 5 + 5 ÷ 5 =6 6+6−6=6 7 − 7 ÷7=6 8 8 8=6 (√9)×(√9) − (√9)=6 complete for 8

Question Number 124944 Answers: 0 Comments: 0

1. (a, m)=(b, m)=1⇒(ab, m)=1 2. c∣ab and (c, a)=1⇒c∣b 3. If c is a common multiple of a and b then [a, b]∣c 4. [ma, mb]=m[a, b] for all int m>0 5. [a, b](a, b)=∣ab∣ 6. Let g>0, s be integers. Show that g∣s iff ∃ integers x, y such that s=x+y and (x, y)=g

$$\mathrm{1}.\:\left({a},\:{m}\right)=\left({b},\:{m}\right)=\mathrm{1}\Rightarrow\left({ab},\:{m}\right)=\mathrm{1} \\ $$$$\mathrm{2}.\:{c}\mid{ab}\:\mathrm{and}\:\left({c},\:{a}\right)=\mathrm{1}\Rightarrow{c}\mid{b} \\ $$$$\mathrm{3}.\:\mathrm{If}\:{c}\:\mathrm{is}\:\mathrm{a}\:\mathrm{common}\:\mathrm{multiple}\:\mathrm{of} \\ $$$${a}\:\mathrm{and}\:{b}\:\mathrm{then}\:\left[{a},\:{b}\right]\mid{c} \\ $$$$\mathrm{4}.\:\left[{ma},\:{mb}\right]={m}\left[{a},\:{b}\right]\:\mathrm{for}\:\mathrm{all}\:\mathrm{int}\:{m}>\mathrm{0} \\ $$$$\mathrm{5}.\:\left[{a},\:{b}\right]\left({a},\:{b}\right)=\mid{ab}\mid \\ $$$$\mathrm{6}.\:\mathrm{Let}\:{g}>\mathrm{0},\:{s}\:\mathrm{be}\:\mathrm{integers}.\:\mathrm{Show} \\ $$$$\mathrm{that}\:{g}\mid{s}\:\mathrm{iff}\:\exists\:\mathrm{integers}\:{x},\:{y}\:\mathrm{such} \\ $$$$\mathrm{that}\:{s}={x}+{y}\:\mathrm{and}\:\left({x},\:{y}\right)={g} \\ $$

Question Number 124795 Answers: 1 Comments: 0

Question Number 124130 Answers: 1 Comments: 0

Let P = ( 3(√6) + 7 )^(89) and F is fractional part of P. Then find the remainder when (PF) + (PF)^2 + (PF)^3 is divided by 31.

$$ \\ $$$$\:\:\mathrm{Let}\:\:\mathrm{P}\:=\:\left(\:\mathrm{3}\sqrt{\mathrm{6}}\:+\:\mathrm{7}\:\right)^{\mathrm{89}} \:\mathrm{and}\:\:\:\mathrm{F}\:\mathrm{is}\:\mathrm{fractional} \\ $$$$\:\:\mathrm{part}\:\mathrm{of}\:\:\mathrm{P}. \\ $$$$\:\:\mathrm{Then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when} \\ $$$$\:\:\:\left(\mathrm{PF}\right)\:+\:\left(\mathrm{PF}\right)^{\mathrm{2}} \:+\:\left(\mathrm{PF}\right)^{\mathrm{3}} \:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{31}. \\ $$

Question Number 122896 Answers: 1 Comments: 0

Prove : (n,7) = 1 → 7∣(n^6 − 1)

$${Prove}\:: \\ $$$$\left({n},\mathrm{7}\right)\:=\:\mathrm{1}\:\rightarrow\:\mathrm{7}\mid\left({n}^{\mathrm{6}} \:−\:\mathrm{1}\right) \\ $$

Question Number 122637 Answers: 0 Comments: 0

can anybody do me a faver and tell me what the direct proof of mobius inversion formula is? (proof without using Dirichlet convolutions)?

$${can}\:{anybody}\:{do}\:{me}\:{a}\:{faver}\:{and}\:{tell}\:{me}\:{what}\:\:{the}\:\:{direct}\:{proof}\:{of}\:\:{mobius}\:{inversion}\:{formula}\:{is}?\:\left({proof}\:{without}\:{using}\:\:{Dirichlet}\:{convolutions}\right)? \\ $$$$ \\ $$

Question Number 121834 Answers: 1 Comments: 0

Question Number 121694 Answers: 1 Comments: 0

Evaluate Σ_(k = 1) ^n ((4k+(√(4k^2 −1)))/( (√(2k−1)) + (√(2k+1)))) ?

$${Evaluate}\:\underset{{k}\:=\:\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{4}{k}+\sqrt{\mathrm{4}{k}^{\mathrm{2}} −\mathrm{1}}}{\:\sqrt{\mathrm{2}{k}−\mathrm{1}}\:+\:\sqrt{\mathrm{2}{k}+\mathrm{1}}}\:? \\ $$

Question Number 120923 Answers: 1 Comments: 0

Determine all primes p for which the system of equations { ((p +1 = 2x^2 )),((p^2 +1=2y^2 )) :} ; has solution in integers x, y.

$$\mathrm{Determine}\:\mathrm{all}\:\mathrm{primes}\:{p}\:\mathrm{for}\:\mathrm{which} \\ $$$$\mathrm{the}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations}\: \\ $$$$\begin{cases}{{p}\:+\mathrm{1}\:=\:\mathrm{2x}^{\mathrm{2}} }\\{{p}^{\mathrm{2}} +\mathrm{1}=\mathrm{2y}^{\mathrm{2}} \:\:}\end{cases}\:;\:\mathrm{has}\:\mathrm{solution}\:\mathrm{in} \\ $$$$\mathrm{integers}\:\mathrm{x},\:\mathrm{y}. \\ $$

Question Number 120812 Answers: 1 Comments: 0

Find all integral solutions to the equation (x^2 +1)(y^2 +1)+2(x−y)(1−xy)=4(1+xy)

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{integral}\:\mathrm{solutions}\:\mathrm{to}\: \\ $$$$\mathrm{the}\:\mathrm{equation}\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{y}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{2}\left(\mathrm{x}−\mathrm{y}\right)\left(\mathrm{1}−\mathrm{xy}\right)=\mathrm{4}\left(\mathrm{1}+\mathrm{xy}\right) \\ $$

Question Number 120679 Answers: 2 Comments: 3

$${h} \\ $$

Question Number 120563 Answers: 1 Comments: 0

A man has a wife with six children and his total income is Gh$8500.He was allowed the following free tax personal:......$1200 wife:...........$300 each child:.....$250 for a maximum of 4 dependant relative:$400 insurance:........$250 Taxed: the first $2000 at 10% the next $2000 at 15% the next $2000 at 20% the next $2000 at 25% calculate a)his tax free income b)his income tax c)his monthly income d)his net monthly income

$${A}\:{man}\:{has}\:{a}\:{wife}\:{with}\:{six} \\ $$$${children}\:{and}\:{his}\:{total}\:{income}\:{is} \\ $$$${Gh\$}\mathrm{8500}.{He}\:{was}\:{allowed}\:{the} \\ $$$${following}\:{free}\:{tax} \\ $$$${personal}:......\$\mathrm{1200} \\ $$$${wife}:...........\$\mathrm{300} \\ $$$${each}\:{child}:.....\$\mathrm{250}\:{for}\:{a}\:{maximum}\:{of}\:\mathrm{4} \\ $$$${dependant}\:{relative}:\$\mathrm{400} \\ $$$${insurance}:........\$\mathrm{250} \\ $$$${Taxed}: \\ $$$${the}\:{first}\:\$\mathrm{2000}\:{at}\:\mathrm{10\%} \\ $$$${the}\:{next}\:\$\mathrm{2000}\:{at}\:\mathrm{15\%} \\ $$$${the}\:{next}\:\$\mathrm{2000}\:{at}\:\mathrm{20\%} \\ $$$${the}\:{next}\:\$\mathrm{2000}\:{at}\:\mathrm{25\%} \\ $$$${calculate} \\ $$$$\left.{a}\right){his}\:{tax}\:{free}\:{income} \\ $$$$\left.{b}\right){his}\:{income}\:{tax} \\ $$$$\left.{c}\right){his}\:{monthly}\:{income} \\ $$$$\left.{d}\right){his}\:{net}\:{monthly}\:{income} \\ $$$$ \\ $$

Question Number 120202 Answers: 0 Comments: 0

Show for the equation a^n = b^2 −1 where n>1 and a>2 are any natural numbers , there are no positive integer solutions for a and b ?

$${Show}\:{for}\:{the}\:{equation}\:{a}^{{n}} \:=\:{b}^{\mathrm{2}} −\mathrm{1}\: \\ $$$${where}\:{n}>\mathrm{1}\:{and}\:{a}>\mathrm{2}\:{are}\:{any}\:{natural} \\ $$$${numbers}\:,\:{there}\:{are}\:{no}\:{positive}\:{integer} \\ $$$${solutions}\:{for}\:{a}\:{and}\:{b}\:? \\ $$

Question Number 120049 Answers: 2 Comments: 0

f(x+2)+f(x−2)=f(x) f(1)=1 ,f(2)=2,f(3)=3,f(4)=4 then f(100)=?

$$\:{f}\left({x}+\mathrm{2}\right)+{f}\left({x}−\mathrm{2}\right)={f}\left({x}\right) \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1}\:,{f}\left(\mathrm{2}\right)=\mathrm{2},{f}\left(\mathrm{3}\right)=\mathrm{3},{f}\left(\mathrm{4}\right)=\mathrm{4} \\ $$$${then}\:{f}\left(\mathrm{100}\right)=? \\ $$

Question Number 119956 Answers: 2 Comments: 0

Given a = 1+3+3^2 +3^3 +3^4 +...+3^(100) Find the remainder of dividing the number by 5 . (a) 2 (b) 0 (c)4 (d)1 (e) 3

$${Given}\:{a}\:=\:\mathrm{1}+\mathrm{3}+\mathrm{3}^{\mathrm{2}} +\mathrm{3}^{\mathrm{3}} +\mathrm{3}^{\mathrm{4}} +...+\mathrm{3}^{\mathrm{100}} \\ $$$${Find}\:{the}\:{remainder}\:{of}\:{dividing}\:{the}\:{number} \\ $$$${by}\:\mathrm{5}\:. \\ $$$$\left({a}\right)\:\mathrm{2}\:\:\:\:\:\left({b}\right)\:\mathrm{0}\:\:\:\:\:\:\:\left({c}\right)\mathrm{4}\:\:\:\:\:\:\left({d}\right)\mathrm{1}\:\:\:\:\:\:\left({e}\right)\:\mathrm{3} \\ $$

Question Number 119790 Answers: 2 Comments: 0

Let x,y,z be nonnegative real numbers, which satisfy x+y+z=1 Find minimum value of Q=(√(2−x)) + (√(2−y)) + (√(2−z)) .

$${Let}\:{x},{y},{z}\:{be}\:{nonnegative}\:{real} \\ $$$${numbers},\:{which}\:{satisfy}\:{x}+{y}+{z}=\mathrm{1} \\ $$$${Find}\:{minimum}\:{value}\:{of}\: \\ $$$${Q}=\sqrt{\mathrm{2}−{x}}\:+\:\sqrt{\mathrm{2}−{y}}\:+\:\sqrt{\mathrm{2}−{z}}\:. \\ $$

Question Number 119657 Answers: 1 Comments: 2

Suppose that the greatest common divisor of the positive integers a,b and c is 1 and ((ab)/(a−b)) = c . Prove that a−b is a perfect square

$${Suppose}\:{that}\:{the}\:{greatest}\:{common}\:{divisor}\:{of} \\ $$$${the}\:{positive}\:{integers}\:{a},{b}\:{and}\:{c}\:{is}\:\mathrm{1}\:{and} \\ $$$$\frac{{ab}}{{a}−{b}}\:=\:{c}\:.\:{Prove}\:{that}\:{a}−{b}\:{is}\:{a} \\ $$$${perfect}\:{square} \\ $$

Question Number 119600 Answers: 2 Comments: 0

Find gcd of x^4 +x^3 −4x^2 +x+5 and x^3 +x^2 −9x−9

$${Find}\:{gcd}\:{of}\:{x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +{x}+\mathrm{5}\: \\ $$$${and}\:{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{9}{x}−\mathrm{9} \\ $$

Question Number 119094 Answers: 1 Comments: 0

Why the Fermat formula about prime numbers is for n = 5 and n = 6 fail? f(n) = 2^2^n + 1

$${Why}\:{the}\:{Fermat}\:{formula}\:{about}\:{prime}\:{numbers}\:{is}\:{for}\:{n}\:=\:\mathrm{5}\:{and}\:{n}\:=\:\mathrm{6}\:{fail}? \\ $$$$ \\ $$$${f}\left({n}\right)\:=\:\mathrm{2}^{\mathrm{2}^{{n}} } \:+\:\mathrm{1} \\ $$

Question Number 118880 Answers: 3 Comments: 0

8(1+(1/2))(1+(1/2^2 ))(1+(1/2^4 ))(1+(1/2^8 ))...(1+(1/2^(32) ))+ (1/2^(60) )=?

$$\mathrm{8}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{8}} }\right)...\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{32}} }\right)+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{60}} }=? \\ $$

Question Number 118357 Answers: 2 Comments: 0

Find the sum of the prime factors of 20^5 +21.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{prime}\:\mathrm{factors}\:\mathrm{of} \\ $$$$\mathrm{20}^{\mathrm{5}} +\mathrm{21}. \\ $$

Question Number 118362 Answers: 1 Comments: 0

Prove that 𝛑 is an irrational number

$$\mathrm{Prove}\:\mathrm{that}\:\boldsymbol{\pi}\:\mathrm{is}\:\mathrm{an}\:\mathrm{irrational}\:\mathrm{number} \\ $$

Question Number 117848 Answers: 0 Comments: 0

Determine all functions f:R→R such that the equality f([x] y)= f(x) [f(y) ] holds for all x,y ∈R . Here by [x] we denote the greatest integer not exceeding x.

$${Determine}\:{all}\:{functions}\:{f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${such}\:{that}\:{the}\:{equality}\:{f}\left(\left[{x}\right]\:{y}\right)=\:{f}\left({x}\right)\:\left[{f}\left({y}\right)\:\right] \\ $$$${holds}\:{for}\:{all}\:{x},{y}\:\in\mathbb{R}\:.\:{Here}\:\:{by}\:\left[{x}\right]\:{we}\: \\ $$$${denote}\:{the}\:{greatest}\:{integer}\:{not}\:{exceeding}\:{x}. \\ $$$$ \\ $$

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