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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

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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?

Guest Feb 4, 2016

#2
+19207
+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

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

heureka  Feb 5, 2016
edited by heureka  Feb 5, 2016
edited by heureka  Feb 5, 2016
Sort:

#1
+19207
+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?

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

heureka  Feb 5, 2016
#2
+19207
+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

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

heureka  Feb 5, 2016
edited by heureka  Feb 5, 2016
edited by heureka  Feb 5, 2016
#3
+85699
+12

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

CPhill  Feb 6, 2016

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