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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!

Toomanyquestionspersinglepost!

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

Q#8 2058000, 2058343, 2058686

You can't use 'macro parameter character #' in math mode2058000,2058343,2058686

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

[V.9.]Lastdigit(−1)×1010+(1)×1010=0Ω10≡010≡0(mod10)⇒Lasttwodigits=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

[V.7.]Ω1−Ω2=(361−251)+(362−252)+...+(36n−25n)36n−25n≡3n−3n≡110⇒Ω1−Ω2≡110

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✓

1a+1+2b+1+3c+1+4d+1=1aa+1+2bb+1+3cc+1+4dd+1=?−.−.−.−.−.−.−.−.−.−.−.−.−.−.−.−▸1a+1+1+2b+1+2+3c+1+3+4d+1+4=1+10▸a+2a+1+2b+4b+1+3c+6c+1+4d+8d+1=11▸(aa+1+2a+1)+(2bb+1+4b+1)+(3cc+1+6c+1)+(4dd+1+8d+1)=11▸(aa+1+2bb+1+3cc+1+4dd+1)+(2a+1+4b+1+6c+1+8d+1)=11▸(aa+1+2bb+1+3cc+1+4dd+1)+2(1a+1+2b+1+3c+1+4d+1)=11▸(aa+1+2bb+1+3cc+1+4dd+1)+2(1)=11▸aa+1+2bb+1+3cc+1+4dd+1=11−2=9✓

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=3,b=7,c=11,d=1534+74+114+154=364=9

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

mnjuly sir, please share your complete solution.

mnjulysir,pleaseshareyourcompletesolution.

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+3a−2b=7ab+3a−2b−6=7−6a(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)

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

letp=aa+1+2bb+1+3cc+1+4dd+1p=a+1−1a+1+2b+2−2b+1+3c+3−3c+1+4d+4−4d+1p=1+2+3+4−(1a+1+2b+1+3c+1+4d+1)⏟1p=10−1=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) .

You can't use 'macro parameter character #' in math modeΩ1+Ω2=676,(Ω1,Ω2)=169LetΩ1=169m&Ω2=169n,where(m,n)=1Ω1+Ω2=169m+169n=676m+n=4∧(m,n)=1(m,n)=(1,3),(3,1)(Ω1,Ω2)=(1×169,3×169),(3×169,1×169)=(169,507),(507,169)Note:Oneofthe2figures(876&169)iswrongatleast.Ihavecorrected:876676.

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

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

Nohopeofanytyperesponsefromthequestioner!!!

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