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Question Number 178115 by aurpeyz last updated on 12/Oct/22

(n/(390)) is a prime number less than 12 and   one of the prime factor of  n is not   a prime factor of 390.  Qty A: n  Qty B: 3000

$$\frac{{n}}{\mathrm{390}}\:{is}\:{a}\:{prime}\:{number}\:{less}\:{than}\:\mathrm{12}\:{and}\: \\ $$$${one}\:{of}\:{the}\:{prime}\:{factor}\:{of}\:\:{n}\:{is}\:{not}\: \\ $$$${a}\:{prime}\:{factor}\:{of}\:\mathrm{390}. \\ $$$${Qty}\:{A}:\:{n} \\ $$$${Qty}\:{B}:\:\mathrm{3000} \\ $$

Answered by Rasheed.Sindhi last updated on 13/Oct/22

(n/(390))=m(∈P)<12  n=390m  let p is a prime factor of n which is  not a prime factor of 390.  let n=pq ∧ p ∣ m   ∵ m∈P          ∴p=m  (n/(390))=p ∧ p∣n but p ∤ 390  ∴p≠1,2,3,5,6,10,13,15,26,30,39,65,            78,130,195,390   p(<12)=7,11  n=390p=390∙7,390∙11  n=2730,4290

$$\frac{{n}}{\mathrm{390}}={m}\left(\in\mathbb{P}\right)<\mathrm{12} \\ $$$${n}=\mathrm{390}{m} \\ $$$${let}\:{p}\:{is}\:{a}\:{prime}\:{factor}\:{of}\:{n}\:{which}\:{is} \\ $$$${not}\:{a}\:{prime}\:{factor}\:{of}\:\mathrm{390}. \\ $$$${let}\:{n}={pq}\:\wedge\:{p}\:\mid\:{m} \\ $$$$\:\because\:{m}\in\mathbb{P}\:\:\:\:\:\:\:\:\:\:\therefore{p}={m} \\ $$$$\frac{{n}}{\mathrm{390}}={p}\:\wedge\:{p}\mid{n}\:{but}\:{p}\:\nmid\:\mathrm{390} \\ $$$$\therefore{p}\neq\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{6},\mathrm{10},\mathrm{13},\mathrm{15},\mathrm{26},\mathrm{30},\mathrm{39},\mathrm{65}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{78},\mathrm{130},\mathrm{195},\mathrm{390}\: \\ $$$${p}\left(<\mathrm{12}\right)=\mathrm{7},\mathrm{11} \\ $$$${n}=\mathrm{390}{p}=\mathrm{390}\centerdot\mathrm{7},\mathrm{390}\centerdot\mathrm{11} \\ $$$${n}=\mathrm{2730},\mathrm{4290} \\ $$

Commented by Tawa11 last updated on 13/Oct/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 13/Oct/22

(n/(390))=(n/(2×3×5×13))=p  p=2,3,5,7,11  ⇒n=2×3×5×13×7=2730 or  ⇒n=2×3×5×13×11=4290

$$\frac{{n}}{\mathrm{390}}=\frac{{n}}{\mathrm{2}×\mathrm{3}×\mathrm{5}×\mathrm{13}}={p} \\ $$$${p}=\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{7},\mathrm{11} \\ $$$$\Rightarrow{n}=\mathrm{2}×\mathrm{3}×\mathrm{5}×\mathrm{13}×\mathrm{7}=\mathrm{2730}\:{or} \\ $$$$\Rightarrow{n}=\mathrm{2}×\mathrm{3}×\mathrm{5}×\mathrm{13}×\mathrm{11}=\mathrm{4290} \\ $$

Commented by Rasheed.Sindhi last updated on 13/Oct/22

A_(SMART_(APPROACH!^( ) ) ^( ) )

$$\underset{\underset{\overset{\:} {\mathrm{APPROACH}!}} {\overset{\:} {\mathrm{SMART}}}} {\mathrm{A}}\:\:\:\: \\ $$

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