1. Rod pRoblems

A carpenter has a long rod of length 4m 38 cm. He needs to cut this rod into lengths of 161cm, 23cm and 7cm so that he would get at least one rod of each length and have nothing left over. After a while, the carpenter managed to solve the problem. How many rods of each length did he get?
Find all possible solutions.

Whoever does this is my heroe

2. Originally Posted by googa
A carpenter has a long rod of length 4m 38 cm. He needs to cut this rod into lengths of 161cm, 23cm and 7cm so that he would get at least one rod of each length and have nothing left over. After a while, the carpenter managed to solve the problem. How many rods of each length did he get?
Find all possible solutions.

Whoever does this is my heroe
Here is one way.

4m 38cm = 438cm

How many 161 cm are in 438 cm?
Two
So, what is left is 438 -2(161) = 116 cm

Let x = number of 23-cm rods
And y = number of 7-cm rods
So,
23x +7y = 116
x = (116 -7y)/23 -----**

By repetition, find the the integer value of y that will produce an integer x.
I found y = 10 to give x = 2
That means 10 rods of 7cm long
And 2 rods of 23 cm long

Therefore, the carpenter got 2 rods of 161cm long, 2 rods of 23cm long, and 10 rods of 7cm long. -------------answer.

Check:
2(161) +2(23) +10(7) =? 438
438 =? 438
Yes, so, OK.

3. let x = number of 161 lengths
let y = number of 23 lengths
let z = number of 7 lengths

Then 161x + 23y + 7z = 438

Note x can only be 1 or 2

if x = 1 then 23y + 7z = 277

substitute all possible values of y (ie 11,10,...,1) . select values which give whole number solution for z.

similarly if x = 2 then 23y + 7z = 116

again substitute all possible y values (ie 4,3,2,1 ). select values which give a whole number solution for z

4. Hello, googa!

Yet another approach . . .

A carpenter has a long rod of length 438 cm.
He needs to cut this rod into lengths of 161cm, 23cm and 7cm
so that he would get at least one rod of each length and have nothing left over.
After a while, the carpenter managed to solve the problem.

How many rods of each length did he get?
Find all possible solutions.
I found three solutions.

Let: .$\displaystyle \begin{array}{ccc}L &=&\text{number of 161-cm rods} \\ M &=& \text{number of 23-cm rods} \\ S &=& \text{number of 7-cm roads} \end{array}$

We have: .$\displaystyle 161L + 23M + 7S \:=\:438 \quad\Rightarrow\quad S \;=\;\frac{438 - 161L - 23M}{7}$

. . Hence: .$\displaystyle S \;=\;62 - 23L - 4M + \frac{4+5M}{7}$

Since $\displaystyle S$ is an integer, $\displaystyle 4 + 5M$ must be divisible by 7.
. . This happens for $\displaystyle M \:=\:2,9,16,\hdots$

If $\displaystyle M = 2$, we have: .$\displaystyle S \:=\:56-23L$ . . . then: .$\displaystyle \begin{array}{cc}L \:=\:1, & S \:=\:33 \\ L \:=\:2, & S \:=\:10 \end{array}$

. Two solutions: .$\displaystyle (L,M,S) \:=\:\boxed{(1,2,33),\;(2,2,10)}$

If $\displaystyle M = 9$, we have: .$\displaystyle S \:=\:33-23L$ . . . then: .$\displaystyle L = 1,\;\;S = 10$

. . One solution: .$\displaystyle (L,M,S) \:=\:\boxed{(1,9,10)}$

If $\displaystyle M = 16$, we have: .$\displaystyle S \:=\:10-23L$ . . . no solutions