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Question-205032

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Question Number 205032 by BaliramKumar last updated on 06/Mar/24
Commented by BaliramKumar last updated on 06/Mar/24
please clarification
$$ \\ $$please clarification
Answered by Rasheed.Sindhi last updated on 06/Mar/24
N=qD+14 where D>14 3N=3qD+42=pD+8 4N=4qD+56=pD+8+qD+14 4qD+56=(p+q)D+22 (p+q)D−4qD=56−22=34 D(p−3q)=34 D_(>14) =((34)/(p−3q)) p−3q=1,2,17,34 D =34,17,2^(×) ,1^(×) ∵ D>14 ∴ D=34,17 Case 1: D=17 N=17q+14≡14(mod 17) 3N=51q+42≡8(mod 17) 4N=4×17q+56≡5 Case 2:D=34 N=34q+14≡14(mod 34) 3N=3×34q+42≡8(mod 34) 4N=4×34q+56≡22(mod 34)
$${N}={qD}+\mathrm{14}\:{where}\:{D}>\mathrm{14} \\ $$$$\mathrm{3}{N}=\mathrm{3}{qD}+\mathrm{42}={pD}+\mathrm{8} \\ $$$$\mathrm{4}{N}=\mathrm{4}{qD}+\mathrm{56}={pD}+\mathrm{8}+{qD}+\mathrm{14} \\ $$$$\mathrm{4}{qD}+\mathrm{56}=\left({p}+{q}\right){D}+\mathrm{22} \\ $$$$\left({p}+{q}\right){D}−\mathrm{4}{qD}=\mathrm{56}−\mathrm{22}=\mathrm{34} \\ $$$${D}\left({p}−\mathrm{3}{q}\right)=\mathrm{34} \\ $$$${D}_{>\mathrm{14}} =\frac{\mathrm{34}}{{p}−\mathrm{3}{q}} \\ $$$$\:{p}−\mathrm{3}{q}=\mathrm{1},\mathrm{2},\mathrm{17},\mathrm{34} \\ $$$${D}\:\:\:\:\:\:\:\:\:=\mathrm{34},\mathrm{17},\overset{×} {\mathrm{2}},\overset{×} {\mathrm{1}} \\ $$$$\because\:{D}>\mathrm{14}\:\:\therefore\:{D}=\mathrm{34},\mathrm{17} \\ $$$${Case}\:\mathrm{1}:\:{D}=\mathrm{17} \\ $$$${N}=\mathrm{17}{q}+\mathrm{14}\equiv\mathrm{14}\left({mod}\:\mathrm{17}\right) \\ $$$$\mathrm{3}{N}=\mathrm{51}{q}+\mathrm{42}\equiv\mathrm{8}\left({mod}\:\mathrm{17}\right) \\ $$$$\mathrm{4}{N}=\mathrm{4}×\mathrm{17}{q}+\mathrm{56}\equiv\mathrm{5} \\ $$$${Case}\:\mathrm{2}:{D}=\mathrm{34} \\ $$$${N}=\mathrm{34}{q}+\mathrm{14}\equiv\mathrm{14}\left({mod}\:\mathrm{34}\right) \\ $$$$\mathrm{3}{N}=\mathrm{3}×\mathrm{34}{q}+\mathrm{42}\equiv\mathrm{8}\left({mod}\:\mathrm{34}\right) \\ $$$$\mathrm{4}{N}=\mathrm{4}×\mathrm{34}{q}+\mathrm{56}\equiv\mathrm{22}\left({mod}\:\mathrm{34}\right) \\ $$
Answered by A5T last updated on 06/Mar/24
N=Dk+14;3N=Dq+8; 4N=Dc+r D≥15 3N=3Dk+14×3=3Dk+34+8 ⇒D∣34 and D≥15 ⇒D=34 or 17 4N=D(k+q)+22; If D=34; then r=22 But if D=17, then 4N=D(k+q+1)+5⇒r=5 ⇒r=5 or r=22
$${N}={Dk}+\mathrm{14};\mathrm{3}{N}={Dq}+\mathrm{8};\:\mathrm{4}{N}={Dc}+{r} \\ $$$${D}\geqslant\mathrm{15} \\ $$$$\mathrm{3}{N}=\mathrm{3}{Dk}+\mathrm{14}×\mathrm{3}=\mathrm{3}{Dk}+\mathrm{34}+\mathrm{8} \\ $$$$\Rightarrow{D}\mid\mathrm{34}\:{and}\:{D}\geqslant\mathrm{15}\:\Rightarrow{D}=\mathrm{34}\:{or}\:\mathrm{17} \\ $$$$\mathrm{4}{N}={D}\left({k}+{q}\right)+\mathrm{22};\:{If}\:{D}=\mathrm{34};\:{then}\:{r}=\mathrm{22} \\ $$$${But}\:{if}\:{D}=\mathrm{17},\:{then}\:\mathrm{4}{N}={D}\left({k}+{q}+\mathrm{1}\right)+\mathrm{5}\Rightarrow{r}=\mathrm{5} \\ $$$$\Rightarrow{r}=\mathrm{5}\:{or}\:{r}=\mathrm{22} \\ $$
Commented by BaliramKumar last updated on 06/Mar/24
Q say proper divisor
$$\mathrm{Q}\:\:\:\:\:\mathrm{say}\:\mathrm{proper}\:\mathrm{divisor} \\ $$
Commented by A5T last updated on 06/Mar/24
The number ,D, cannot be a proper “divisor”, otherwise it′d divide N completely without a remainder.
$${The}\:{number}\:,{D},\:{cannot}\:{be}\:{a}\:{proper}\:“{divisor}'', \\ $$$${otherwise}\:{it}'{d}\:{divide}\:{N}\:{completely}\:{without}\:{a}\: \\ $$$${remainder}. \\ $$
Commented by BaliramKumar last updated on 06/Mar/24
thanks
$$\mathrm{thanks} \\ $$
Answered by Rasheed.Sindhi last updated on 06/Mar/24
N≡14(mod D)⇒D>14 3N≡42≡8(mod D) D∣(42−8)⇒D∣34 ∧ D>14 D=17,34 N=17q_1 +14 , 34q_1 +14 Case1:D=17 N=17q_1 +14≡14(mod 17) 4N=4×17q_1 +56≡56(mod 17) 4N≡56−3×17=5(mod 17 Case1:D=34 N=34q_2 +14≡14(mod 34) 4N=4×34q_1 +56≡56(mod 34) 4N≡56−34=22(mod 34)
$${N}\equiv\mathrm{14}\left({mod}\:{D}\right)\Rightarrow{D}>\mathrm{14} \\ $$$$\mathrm{3}{N}\equiv\mathrm{42}\equiv\mathrm{8}\left({mod}\:{D}\right) \\ $$$${D}\mid\left(\mathrm{42}−\mathrm{8}\right)\Rightarrow{D}\mid\mathrm{34}\:\wedge\:{D}>\mathrm{14} \\ $$$${D}=\mathrm{17},\mathrm{34} \\ $$$${N}=\mathrm{17}{q}_{\mathrm{1}} +\mathrm{14}\:,\:\mathrm{34}{q}_{\mathrm{1}} +\mathrm{14} \\ $$$${Case}\mathrm{1}:{D}=\mathrm{17} \\ $$$${N}=\mathrm{17}{q}_{\mathrm{1}} +\mathrm{14}\equiv\mathrm{14}\left({mod}\:\mathrm{17}\right) \\ $$$$\mathrm{4}{N}=\mathrm{4}×\mathrm{17}{q}_{\mathrm{1}} +\mathrm{56}\equiv\mathrm{56}\left({mod}\:\mathrm{17}\right) \\ $$$$\mathrm{4}{N}\equiv\mathrm{56}−\mathrm{3}×\mathrm{17}=\mathrm{5}\left({mod}\:\mathrm{17}\right. \\ $$$${Case}\mathrm{1}:{D}=\mathrm{34} \\ $$$${N}=\mathrm{34}{q}_{\mathrm{2}} +\mathrm{14}\equiv\mathrm{14}\left({mod}\:\mathrm{34}\right) \\ $$$$\mathrm{4}{N}=\mathrm{4}×\mathrm{34}{q}_{\mathrm{1}} +\mathrm{56}\equiv\mathrm{56}\left({mod}\:\mathrm{34}\right) \\ $$$$\mathrm{4}{N}\equiv\mathrm{56}−\mathrm{34}=\mathrm{22}\left({mod}\:\mathrm{34}\right) \\ $$
Commented by BaliramKumar last updated on 06/Mar/24
thanks
$$\mathrm{thanks} \\ $$

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