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Question Number 113508 by Lekhraj last updated on 13/Sep/20

Commented by mr W last updated on 13/Sep/20

i got N=LCM(5,7,8,9)=2520

$${i}\:{got}\:{N}={LCM}\left(\mathrm{5},\mathrm{7},\mathrm{8},\mathrm{9}\right)=\mathrm{2520} \\ $$

Commented by Rasheed.Sindhi last updated on 13/Sep/20

                   N=252

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{N}=\mathrm{252} \\ $$

Answered by Rasheed.Sindhi last updated on 13/Sep/20

10N+1≡0(mod 1)  No restriction on N  10N+2≡0(mod 2)  No restriction on N  10N+3≡0(mod 3)  ⇒3 ∣ N  10N+4≡0(mod 4)   ⇒2 ∣ N  10N+5≡0(mod 5)  No restriction on N  10N+6≡0(mod 6)  ⇒3 ∣ N  10N+7≡0(mod 7)  ⇒7 ∣ N  10N+8≡0(mod 8)  ⇒4 ∣ N  10N+9≡0(mod 9)  ⇒9 ∣ N   Hence N is the least number  which is divisible by 2,3,4,7 & 9  N=LCM(2,3,4,7,9)       =LCM(4,7,9)=252

$$\mathrm{10N}+\mathrm{1}\equiv\mathrm{0}\left({mod}\:\mathrm{1}\right) \\ $$$${No}\:{restriction}\:{on}\:\mathrm{N} \\ $$$$\mathrm{10N}+\mathrm{2}\equiv\mathrm{0}\left({mod}\:\mathrm{2}\right) \\ $$$${No}\:{restriction}\:{on}\:\mathrm{N} \\ $$$$\mathrm{10N}+\mathrm{3}\equiv\mathrm{0}\left({mod}\:\mathrm{3}\right) \\ $$$$\Rightarrow\mathrm{3}\:\mid\:\mathrm{N} \\ $$$$\mathrm{10N}+\mathrm{4}\equiv\mathrm{0}\left({mod}\:\mathrm{4}\right) \\ $$$$\:\Rightarrow\mathrm{2}\:\mid\:\mathrm{N} \\ $$$$\mathrm{10N}+\mathrm{5}\equiv\mathrm{0}\left({mod}\:\mathrm{5}\right) \\ $$$${No}\:{restriction}\:{on}\:\mathrm{N} \\ $$$$\mathrm{10N}+\mathrm{6}\equiv\mathrm{0}\left({mod}\:\mathrm{6}\right) \\ $$$$\Rightarrow\mathrm{3}\:\mid\:\mathrm{N} \\ $$$$\mathrm{10N}+\mathrm{7}\equiv\mathrm{0}\left({mod}\:\mathrm{7}\right) \\ $$$$\Rightarrow\mathrm{7}\:\mid\:\mathrm{N} \\ $$$$\mathrm{10N}+\mathrm{8}\equiv\mathrm{0}\left({mod}\:\mathrm{8}\right) \\ $$$$\Rightarrow\mathrm{4}\:\mid\:\mathrm{N} \\ $$$$\mathrm{10N}+\mathrm{9}\equiv\mathrm{0}\left({mod}\:\mathrm{9}\right) \\ $$$$\Rightarrow\mathrm{9}\:\mid\:\mathrm{N}\: \\ $$$${Hence}\:{N}\:{is}\:{the}\:{least}\:{number} \\ $$$${which}\:{is}\:{divisible}\:{by}\:\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{7}\:\&\:\mathrm{9} \\ $$$$\mathrm{N}=\mathrm{LCM}\left(\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{7},\mathrm{9}\right) \\ $$$$\:\:\:\:\:=\mathrm{LCM}\left(\mathrm{4},\mathrm{7},\mathrm{9}\right)=\mathrm{252} \\ $$

Commented by mr W last updated on 13/Sep/20

nice solution sir!

$${nice}\:{solution}\:{sir}! \\ $$

Commented by Rasheed.Sindhi last updated on 14/Sep/20

Thanks sir!

$$\mathcal{T}{hanks}\:\boldsymbol{{sir}}! \\ $$

Commented by Lekhraj last updated on 14/Sep/20

Thanks sir for correction

$${T}\mathrm{hanks}\:\mathrm{sir}\:\mathrm{for}\:\mathrm{correction} \\ $$

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