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Question Number 196659 by sonukgindia last updated on 29/Aug/23

Commented by Rasheed.Sindhi last updated on 29/Aug/23

Too many questions per single post!

$${Too}\:{many}\:{questions}\:{per}\:{single}\:{post}! \\ $$

Commented by Rasheed.Sindhi last updated on 29/Aug/23

Q#8 2058000, 2058343, 2058686

$${Q}#\mathrm{8} \\ $$$$\mathrm{2058000},\:\mathrm{2058343},\:\mathrm{2058686} \\ $$

Commented by deleteduser1 last updated on 29/Aug/23

[V.9.]Last digit (−1)×1010+(1)×1010=0 (Ω/(10))≡(0/(10))≡0(mod 10)⇒Last two digits=00

$$\left[{V}.\mathrm{9}.\right]{Last}\:{digit}\:\left(−\mathrm{1}\right)×\mathrm{1010}+\left(\mathrm{1}\right)×\mathrm{1010}=\mathrm{0} \\ $$$$\frac{\Omega}{\mathrm{10}}\equiv\frac{\mathrm{0}}{\mathrm{10}}\equiv\mathrm{0}\left({mod}\:\mathrm{10}\right)\Rightarrow{Last}\:{two}\:{digits}=\mathrm{00} \\ $$

Commented by deleteduser1 last updated on 29/Aug/23

[V.7.] Ω_1 −Ω_2 =(36^1 −25^1 )+(36^2 −25^2 )+...+(36^n −25^n ) 36^n −25^n ≡3^n −3^n ≡^(11) 0⇒Ω_1 −Ω_2 ≡^(11) 0

$$\left[{V}.\mathrm{7}.\right]\:\Omega_{\mathrm{1}} −\Omega_{\mathrm{2}} =\left(\mathrm{36}^{\mathrm{1}} −\mathrm{25}^{\mathrm{1}} \right)+\left(\mathrm{36}^{\mathrm{2}} −\mathrm{25}^{\mathrm{2}} \right)+...+\left(\mathrm{36}^{{n}} −\mathrm{25}^{{n}} \right) \\ $$$$\mathrm{36}^{{n}} −\mathrm{25}^{{n}} \equiv\mathrm{3}^{{n}} −\mathrm{3}^{{n}} \overset{\mathrm{11}} {\equiv}\mathrm{0}\Rightarrow\Omega_{\mathrm{1}} −\Omega_{\mathrm{2}} \overset{\mathrm{11}} {\equiv}\mathrm{0} \\ $$

Answered by Rasheed.Sindhi last updated on 29/Aug/23

(1/(a+1))+(2/(b+1))+(3/(c+1))+(4/(d+1))=1 (a/(a+1))+((2b)/(b+1))+((3c)/(c+1))+((4d)/(d+1))=? −.−.−.−.−.−.−.−.−.−.−.−.−.−.−.− ▶(1/(a+1))+1+(2/(b+1))+2+(3/(c+1))+3+(4/(d+1))+4=1+10 ▶((a+2)/(a+1))+((2b+4)/(b+1))+((3c+6)/(c+1))+((4d+8)/(d+1))=11 ▶((a/(a+1))+(2/(a+1)))+(((2b)/(b+1))+(4/(b+1))) +(((3c)/(c+1))+(6/(c+1)))+(((4d)/(d+1))+(8/(d+1)))=11 ▶((a/(a+1))+((2b)/(b+1))+((3c)/(c+1))+((4d)/(d+1))) +((2/(a+1))+(4/(b+1))+(6/(c+1))+(8/(d+1)))=11 ▶((a/(a+1))+((2b)/(b+1))+((3c)/(c+1))+((4d)/(d+1))) +2((1/(a+1))+(2/(b+1))+(3/(c+1))+(4/(d+1)))=11 ▶((a/(a+1))+((2b)/(b+1))+((3c)/(c+1))+((4d)/(d+1))) +2(1)=11 ▶(a/(a+1))+((2b)/(b+1))+((3c)/(c+1))+((4d)/(d+1))=11−2=9✓

$$\frac{\mathrm{1}}{{a}+\mathrm{1}}+\frac{\mathrm{2}}{{b}+\mathrm{1}}+\frac{\mathrm{3}}{{c}+\mathrm{1}}+\frac{\mathrm{4}}{{d}+\mathrm{1}}=\mathrm{1} \\ $$$$\frac{{a}}{{a}+\mathrm{1}}+\frac{\mathrm{2}{b}}{{b}+\mathrm{1}}+\frac{\mathrm{3}{c}}{{c}+\mathrm{1}}+\frac{\mathrm{4}{d}}{{d}+\mathrm{1}}=? \\ $$$$−.−.−.−.−.−.−.−.−.−.−.−.−.−.−.− \\ $$$$\blacktriangleright\frac{\mathrm{1}}{{a}+\mathrm{1}}+\mathrm{1}+\frac{\mathrm{2}}{{b}+\mathrm{1}}+\mathrm{2}+\frac{\mathrm{3}}{{c}+\mathrm{1}}+\mathrm{3}+\frac{\mathrm{4}}{{d}+\mathrm{1}}+\mathrm{4}=\mathrm{1}+\mathrm{10} \\ $$$$\blacktriangleright\frac{{a}+\mathrm{2}}{{a}+\mathrm{1}}+\frac{\mathrm{2}{b}+\mathrm{4}}{{b}+\mathrm{1}}+\frac{\mathrm{3}{c}+\mathrm{6}}{{c}+\mathrm{1}}+\frac{\mathrm{4}{d}+\mathrm{8}}{{d}+\mathrm{1}}=\mathrm{11} \\ $$$$\blacktriangleright\left(\frac{{a}}{{a}+\mathrm{1}}+\frac{\mathrm{2}}{{a}+\mathrm{1}}\right)+\left(\frac{\mathrm{2}{b}}{{b}+\mathrm{1}}+\frac{\mathrm{4}}{{b}+\mathrm{1}}\right) \\ $$$$\:\:\:\:\:+\left(\frac{\mathrm{3}{c}}{{c}+\mathrm{1}}+\frac{\mathrm{6}}{{c}+\mathrm{1}}\right)+\left(\frac{\mathrm{4}{d}}{{d}+\mathrm{1}}+\frac{\mathrm{8}}{{d}+\mathrm{1}}\right)=\mathrm{11} \\ $$$$\blacktriangleright\left(\frac{{a}}{{a}+\mathrm{1}}+\frac{\mathrm{2}{b}}{{b}+\mathrm{1}}+\frac{\mathrm{3}{c}}{{c}+\mathrm{1}}+\frac{\mathrm{4}{d}}{{d}+\mathrm{1}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:+\left(\frac{\mathrm{2}}{{a}+\mathrm{1}}+\frac{\mathrm{4}}{{b}+\mathrm{1}}+\frac{\mathrm{6}}{{c}+\mathrm{1}}+\frac{\mathrm{8}}{{d}+\mathrm{1}}\right)=\mathrm{11} \\ $$$$\blacktriangleright\left(\frac{{a}}{{a}+\mathrm{1}}+\frac{\mathrm{2}{b}}{{b}+\mathrm{1}}+\frac{\mathrm{3}{c}}{{c}+\mathrm{1}}+\frac{\mathrm{4}{d}}{{d}+\mathrm{1}}\right) \\ $$$$\:\:\:\:\:\:\:+\mathrm{2}\left(\frac{\mathrm{1}}{{a}+\mathrm{1}}+\frac{\mathrm{2}}{{b}+\mathrm{1}}+\frac{\mathrm{3}}{{c}+\mathrm{1}}+\frac{\mathrm{4}}{{d}+\mathrm{1}}\right)=\mathrm{11} \\ $$$$\blacktriangleright\left(\frac{{a}}{{a}+\mathrm{1}}+\frac{\mathrm{2}{b}}{{b}+\mathrm{1}}+\frac{\mathrm{3}{c}}{{c}+\mathrm{1}}+\frac{\mathrm{4}{d}}{{d}+\mathrm{1}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2}\left(\mathrm{1}\right)=\mathrm{11} \\ $$$$\blacktriangleright\frac{{a}}{{a}+\mathrm{1}}+\frac{\mathrm{2}{b}}{{b}+\mathrm{1}}+\frac{\mathrm{3}{c}}{{c}+\mathrm{1}}+\frac{\mathrm{4}{d}}{{d}+\mathrm{1}}=\mathrm{11}−\mathrm{2}=\mathrm{9}\checkmark \\ $$$$\: \\ $$

Commented by mnjuly1970 last updated on 29/Aug/23

a=3 , b=7 , c=11,d=15 (3/(4 )) + (7/4) +((11)/4) +((15)/4) = ((36)/4) =9

$${a}=\mathrm{3}\:\:,\:{b}=\mathrm{7}\:,\:{c}=\mathrm{11},{d}=\mathrm{15} \\ $$$$\:\:\frac{\mathrm{3}}{\mathrm{4}\:}\:+\:\frac{\mathrm{7}}{\mathrm{4}}\:+\frac{\mathrm{11}}{\mathrm{4}}\:+\frac{\mathrm{15}}{\mathrm{4}}\:=\:\frac{\mathrm{36}}{\mathrm{4}}\:=\mathrm{9} \\ $$

Commented by Rasheed.Sindhi last updated on 29/Aug/23

mnjuly sir, please share your complete solution.

$${mnjuly}\:{sir},\:{please}\:{share}\:{your} \\ $$$$\:{complete}\:{solution}. \\ $$

Answered by Rasheed.Sindhi last updated on 29/Aug/23

ab+3a−2b=7 ab+3a−2b−6=7−6 a(b+3)−2(b+3)=1 (a−2)(b+3)=1=1×1=−1×−1 { ((a−2=1 ∧ b+3=1⇒(a,b)=(3,−2))),((a−2=−1 ∧ b+3=−1⇒(a,b)=(1,−4))) :}

$${ab}+\mathrm{3}{a}−\mathrm{2}{b}=\mathrm{7} \\ $$$${ab}+\mathrm{3}{a}−\mathrm{2}{b}−\mathrm{6}=\mathrm{7}−\mathrm{6} \\ $$$${a}\left({b}+\mathrm{3}\right)−\mathrm{2}\left({b}+\mathrm{3}\right)=\mathrm{1} \\ $$$$\left({a}−\mathrm{2}\right)\left({b}+\mathrm{3}\right)=\mathrm{1}=\mathrm{1}×\mathrm{1}=−\mathrm{1}×−\mathrm{1} \\ $$$$\begin{cases}{{a}−\mathrm{2}=\mathrm{1}\:\wedge\:{b}+\mathrm{3}=\mathrm{1}\Rightarrow\left({a},{b}\right)=\left(\mathrm{3},−\mathrm{2}\right)}\\{{a}−\mathrm{2}=−\mathrm{1}\:\wedge\:{b}+\mathrm{3}=−\mathrm{1}\Rightarrow\left({a},{b}\right)=\left(\mathrm{1},−\mathrm{4}\right)}\end{cases} \\ $$

Answered by universe last updated on 29/Aug/23

let p = (a/(a+1))+((2b)/(b+1))+((3c)/(c+1))+((4d)/(d+1)) p = ((a+1−1)/(a+1))+((2b+2−2)/(b+1))+((3c+3−3)/(c+1))+((4d+4−4)/(d+1)) p = 1+2+3+4−((1/(a+1))+(2/(b+1))+(3/(c+1))+(4/(d+1)))_(1) p = 10−1= 9

$${let}\:{p}\:=\:\frac{{a}}{{a}+\mathrm{1}}+\frac{\mathrm{2}{b}}{{b}+\mathrm{1}}+\frac{\mathrm{3}{c}}{{c}+\mathrm{1}}+\frac{\mathrm{4}{d}}{{d}+\mathrm{1}} \\ $$$$\:\:{p}\:\:=\:\frac{{a}+\mathrm{1}−\mathrm{1}}{{a}+\mathrm{1}}+\frac{\mathrm{2}{b}+\mathrm{2}−\mathrm{2}}{{b}+\mathrm{1}}+\frac{\mathrm{3}{c}+\mathrm{3}−\mathrm{3}}{{c}+\mathrm{1}}+\frac{\mathrm{4}{d}+\mathrm{4}−\mathrm{4}}{{d}+\mathrm{1}} \\ $$$${p}\:=\:\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}−\underset{\mathrm{1}} {\underbrace{\left(\frac{\mathrm{1}}{{a}+\mathrm{1}}+\frac{\mathrm{2}}{{b}+\mathrm{1}}+\frac{\mathrm{3}}{{c}+\mathrm{1}}+\frac{\mathrm{4}}{{d}+\mathrm{1}}\right)}} \\ $$$${p}\:=\:\mathrm{10}−\mathrm{1}=\:\mathrm{9} \\ $$

Answered by Rasheed.Sindhi last updated on 29/Aug/23

Q#6 Ω_1 +Ω_2 =676 ,(Ω_1 ,Ω_2 )=169 Let Ω_1 =169m & Ω_2 =169n, where (m,n)=1 Ω_1 +Ω_2 =169m+169n=676 m+n=4 ∧ (m,n)=1 (m,n)=(1,3),(3,1) (Ω_1 ,Ω_2 )=(1×169,3×169),(3×169,1×169) =(169,507),(507,169) Note:One of the 2 figures(876 & 169) is wrong at least. I have corrected: 876^(676) .

$${Q}#\mathrm{6}\: \\ $$$$\Omega_{\mathrm{1}} +\Omega_{\mathrm{2}} =\mathrm{676}\:,\left(\Omega_{\mathrm{1}} ,\Omega_{\mathrm{2}} \right)=\mathrm{169} \\ $$$${Let}\:\Omega_{\mathrm{1}} =\mathrm{169}{m}\:\&\:\Omega_{\mathrm{2}} =\mathrm{169}{n},\:{where} \\ $$$$\left({m},{n}\right)=\mathrm{1} \\ $$$$\Omega_{\mathrm{1}} +\Omega_{\mathrm{2}} =\mathrm{169}{m}+\mathrm{169}{n}=\mathrm{676} \\ $$$${m}+{n}=\mathrm{4}\:\wedge\:\left({m},{n}\right)=\mathrm{1} \\ $$$$\left({m},{n}\right)=\left(\mathrm{1},\mathrm{3}\right),\left(\mathrm{3},\mathrm{1}\right) \\ $$$$\left(\Omega_{\mathrm{1}} ,\Omega_{\mathrm{2}} \right)=\left(\mathrm{1}×\mathrm{169},\mathrm{3}×\mathrm{169}\right),\left(\mathrm{3}×\mathrm{169},\mathrm{1}×\mathrm{169}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{169},\mathrm{507}\right),\left(\mathrm{507},\mathrm{169}\right) \\ $$$${Note}:{One}\:{of}\:{the}\:\mathrm{2}\:{figures}\left(\mathrm{876}\:\&\:\mathrm{169}\right)\:{is}\:{wrong}\:{at}\:{least}. \\ $$$${I}\:{have}\:{corrected}:\:\cancel{\overset{\mathrm{676}} {\mathrm{876}}}\:. \\ $$$$ \\ $$

Commented by Rasheed.Sindhi last updated on 29/Aug/23

No hope of any type response from the questioner!!!

$$\mathcal{N}{o}\:{hope}\:{of}\:{any}\:{type}\:{response}\:{from} \\ $$$${the}\:{questioner}!!! \\ $$

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