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Question Number 133757 by liberty last updated on 24/Feb/21

	Answered by bobhans last updated on 24/Feb/21
	$Let x be the least number of marbles in the box , such that { ((x≡5 (mod 7))),((x≡6 (mod 11))),((x≡ 8 (mod 13))) :} We can use Chinese remainder theorem a_1 =5 ; a_2 =6 ; a_3 =8 m_1 =7 ,m_2 =11 ,m_3 =13 ⇒ m=7×11×13=1001 determinant (((M_1 =(m/m_1 )=((1001)/7)=143)),((M_2 =(m/m_2 )=((1001)/(11))=91)),((M_3 =(m/m_3 )=((1001)/(13))=77))) ⇒M_1 y_1 ≡1 (mod 7)⇒143y_1 ≡1 (mod 7) y_1 ≡ 5 (mod 7) M_2 y_2 ≡1 (mod 11)⇒91y_2 ≡1 (mod 11) y_2 ≡4 (mod 11) M_3 y_3 ≡1 (mod 13)⇒77y_3 ≡1 (mod 13) y_3 ≡12 (mod 13) Then x ≡ M_1 y_1 a_1 +M_2 y_2 a_2 +M_3 y_3 a_3 (mod 1001) x≡ 143×5×5 + 91×4×6+ 77×12×8 (mod 1001) x≡ 138 (mod 1001)$
	$L e t x b e t h e l e a s t n u m b e r o f m a r b l e s$ $i n t h e b o x, s u c h t h a t {\begin{cases} x \equiv 5 (m o d 7) \\ x \equiv 6 (m o d 11) \\ x \equiv 8 (m o d 13) \end{cases}$ $W e c a n u s e C h i n e s e r e m a i n d e r t h e o r e m$ $a_{1} = 5; a_{2} = 6; a_{3} = 8$ $m_{1} = 7, m_{2} = 11, m_{3} = 13 \Rightarrow m = 7 \times 11 \times 13 = 1001$ $\begin{array}{ccc} M_{1} = \frac{m}{m_{1}} = \frac{1001}{7} = 143 \\ M_{2} = \frac{m}{m_{2}} = \frac{1001}{11} = 91 \\ M_{3} = \frac{m}{m_{3}} = \frac{1001}{13} = 77 \end{array}$ $\Rightarrow M_{1} y_{1} \equiv 1 (m o d 7) \Rightarrow 143 y_{1} \equiv 1 (m o d 7)$ $y_{1} \equiv 5 (m o d 7)$ $M_{2} y_{2} \equiv 1 (m o d 11) \Rightarrow 91 y_{2} \equiv 1 (m o d 11)$ $y_{2} \equiv 4 (m o d 11)$ $M_{3} y_{3} \equiv 1 (m o d 13) \Rightarrow 77 y_{3} \equiv 1 (m o d 13)$ $y_{3} \equiv 12 (m o d 13)$ $T h e n x \equiv M_{1} y_{1} a_{1} + M_{2} y_{2} a_{2} + M_{3} y_{3} a_{3} (m o d 1001)$ $x \equiv 143 \times 5 \times 5 + 91 \times 4 \times 6 + 77 \times 12 \times 8 (m o d 1001)$ $x \equiv 138 (m o d 1001)$
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