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Imagine you have a large supply of 3kg and 8kg weights. Two 3kg weights and three 8kg weights have a mean of 6kg, because (3+3+8+8+8)/5=6 Can you find other combinations of 3kg and 8kg weights whose mean weight is is whole number of kg? What's the smallest? Whats the largest? Can you make all the whole numbers inbetween?cheeky

 Feb 4, 2016

Best Answer 

 #2
avatar+23301 
+35

Imagine you have a large supply of 3kg and 8kg weights. Two 3kg weights and three 8kg weights have a mean of 6kg, because (3+3+8+8+8)/5=6 Can you find other combinations of 3kg and 8kg weights whose mean weight is is whole number of kg? What's the smallest? Whats the largest? Can you make all the whole numbers inbetween?

 

Part 2:

 

\(\text{proof}:\\ \begin{array}{rcll} \text{mean } &=& \frac{ u\cdot n \times 3\ kg + v\cdot n \times 8\ kg } { u\cdot n + v\cdot n } \\ &=& \frac{ n\cdot ( u \times 3\ kg + v \times 8\ kg ) } { n\cdot ( u + v ) } \\ &=& \frac{ u \times 3\ kg + v \times 8\ kg } { u + v } \qquad & | \qquad 3\ kg = 8\ kg - 5\ kg \\ &=& \frac{u \times (8\ kg - 5\ kg) + v \times 8\ kg }{ u + v } \\ &=& \frac{u \times (8 - 5 ) + v \times 8} { u + v} \\ &=& \frac{ 8u - 5u + 8v }{ u + v} \\ &=& \frac{ 8(u+v) - 5u}{ u + v} \\ &=& \frac{ 8(u+v) }{ u + v} - \frac{ 5u }{ u + v} \\ &=& 8 - \frac{ 5u }{ u + v} \\ \end{array}\)

 

mean is a whole number, if \(\frac{ 5u }{ u + v}\) is a whole number.

The divisors of 5 is 5 and 1.

So \(\frac{ 5u }{ u + v}\) is a whole number, if \(u+v= 1\) or \(u+v = 5\)

 

1.)  u+v= 1

\(\begin{array}{|rcrcc|ccc|} \hline u && v && u+v & 8 - \frac{ 5u }{ u + v}&& mean \\ \hline 1 &+& 0 & = & 1 & 8 - \frac{ 5\cdot 1 }{ 1 + 0} &=& 3 \\ 0 &+& 1 & = & 1 & 8 - \frac{ 5\cdot 0 }{ 0 + 1} &=& 8 \\ \hline \end{array}\)

 

2.)  u+v= 5

\(\begin{array}{|rcrcc|ccc|} \hline u && v && u+v & 8 - \frac{ 5u }{ u + v}&& mean \\ \hline 1 &+& 4 & = & 5 & 8 - \frac{ 5\cdot 1 }{ 1 + 4} &=& 7 \\ 2 &+& 3 & = & 5 & 8 - \frac{ 5\cdot 2 }{ 2 + 3} &=& 6 \\ 3 &+& 2 & = & 5 & 8 - \frac{ 5\cdot 3 }{ 3 + 2} &=& 5 \\ 4 &+& 1 & = & 5 & 8 - \frac{ 5\cdot 4 }{ 4 + 1} &=& 4 \\ \hline \end{array}\)

 

laugh

 Feb 5, 2016
edited by heureka  Feb 5, 2016
edited by heureka  Feb 5, 2016
 #1
avatar+23301 
+35

Imagine you have a large supply of 3kg and 8kg weights.

Two 3kg weights and three 8kg weights have a mean of 6kg, because (3+3+8+8+8)/5=6 .

Can you find other combinations of 3kg and 8kg weights whose mean weight is whole number of kg?

What's the smallest?

Whats the largest?

Can you make all the whole numbers inbetween?

 

\(\begin{array}{|r|r|r|r|} \hline \text{mean} & \text{all combinations} & \text{example combination }~ (n= 1) & \text{example mean }~ (n= 1) \\ \hline 3 & 1\times 3\ kg & \\ 4 & 4\cdot n\times 3\ kg + n\times 8\ kg & 4\cdot 1\times 3\ kg + 1\times 8\ kg & \frac{12+8}{4+1} = 4 \\ 5 & 3\cdot n\times 3\ kg + 2\cdot n\times 8\ kg & 3\cdot 1\times 3\ kg + 2\cdot 1\times 8\ kg & \frac{9+16}{3+2} = 5 \\ 6 & 2\cdot n\times 3\ kg + 3\cdot n\times 8\ kg & 2\cdot 1\times 3\ kg + 3\cdot 1\times 8\ kg & \frac{6+24}{2+3} = 6 \\ 7 & n\times 3\ kg + 4\cdot n\times 8\ kg & 1\times 3\ kg + 4\cdot 1\times 8\ kg & \frac{3+32}{1+4} = 7 \\ 8 & 1\times 8\ kg & \\ \hline \end{array}\)

 

laugh

 Feb 5, 2016
 #2
avatar+23301 
+35
Best Answer

Imagine you have a large supply of 3kg and 8kg weights. Two 3kg weights and three 8kg weights have a mean of 6kg, because (3+3+8+8+8)/5=6 Can you find other combinations of 3kg and 8kg weights whose mean weight is is whole number of kg? What's the smallest? Whats the largest? Can you make all the whole numbers inbetween?

 

Part 2:

 

\(\text{proof}:\\ \begin{array}{rcll} \text{mean } &=& \frac{ u\cdot n \times 3\ kg + v\cdot n \times 8\ kg } { u\cdot n + v\cdot n } \\ &=& \frac{ n\cdot ( u \times 3\ kg + v \times 8\ kg ) } { n\cdot ( u + v ) } \\ &=& \frac{ u \times 3\ kg + v \times 8\ kg } { u + v } \qquad & | \qquad 3\ kg = 8\ kg - 5\ kg \\ &=& \frac{u \times (8\ kg - 5\ kg) + v \times 8\ kg }{ u + v } \\ &=& \frac{u \times (8 - 5 ) + v \times 8} { u + v} \\ &=& \frac{ 8u - 5u + 8v }{ u + v} \\ &=& \frac{ 8(u+v) - 5u}{ u + v} \\ &=& \frac{ 8(u+v) }{ u + v} - \frac{ 5u }{ u + v} \\ &=& 8 - \frac{ 5u }{ u + v} \\ \end{array}\)

 

mean is a whole number, if \(\frac{ 5u }{ u + v}\) is a whole number.

The divisors of 5 is 5 and 1.

So \(\frac{ 5u }{ u + v}\) is a whole number, if \(u+v= 1\) or \(u+v = 5\)

 

1.)  u+v= 1

\(\begin{array}{|rcrcc|ccc|} \hline u && v && u+v & 8 - \frac{ 5u }{ u + v}&& mean \\ \hline 1 &+& 0 & = & 1 & 8 - \frac{ 5\cdot 1 }{ 1 + 0} &=& 3 \\ 0 &+& 1 & = & 1 & 8 - \frac{ 5\cdot 0 }{ 0 + 1} &=& 8 \\ \hline \end{array}\)

 

2.)  u+v= 5

\(\begin{array}{|rcrcc|ccc|} \hline u && v && u+v & 8 - \frac{ 5u }{ u + v}&& mean \\ \hline 1 &+& 4 & = & 5 & 8 - \frac{ 5\cdot 1 }{ 1 + 4} &=& 7 \\ 2 &+& 3 & = & 5 & 8 - \frac{ 5\cdot 2 }{ 2 + 3} &=& 6 \\ 3 &+& 2 & = & 5 & 8 - \frac{ 5\cdot 3 }{ 3 + 2} &=& 5 \\ 4 &+& 1 & = & 5 & 8 - \frac{ 5\cdot 4 }{ 4 + 1} &=& 4 \\ \hline \end{array}\)

 

laugh

heureka Feb 5, 2016
edited by heureka  Feb 5, 2016
edited by heureka  Feb 5, 2016
 #3
avatar+104793 
+12

Very nice, heureka.....particularly, the proof part.....!!!!

 

 

 

cool cool cool

 Feb 6, 2016

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