Help please urgent...???

Question

A carpenter had a long rod of length 5m 83cm. He needs to cut this rod into lengths of 161cm, 23cm and 7 cm so that he would get at leats one rod of each length and have nothing left over. After a while, the carpenter managed to solve this question. How many rods of each length did he get? Find all possible solutions.

Answer 1

It's should be solved by trial and error. Let me help you.

Let x, y, z be the number of 161-rod, 23-rod and 7-rod respectively.

161x + 23y + 7z = 583

0 + 2y + 0 = 2 (mod 7)

y = 1 (mod 7)

So, y = 1, y = 8, y = 15, or y = 22

for y = 1

--->161x + 7z = 560

---> x = 3, z = 11

---> x = 2, z = 34

---> x = 1, z = 57

for y = 8

--->161x + 7z = 399

--->x = 2, z = 11

--->x = 1, z = 34

for y = 15

--->161x + 7z = 238

--->x = 1, z = 11

for y = 22

--->161x + 7z = 77

--->x = 0 (not a solution)

So, all the solutions are: (1,15,11), (1,8,34), (2,8,11), (1,1,57), (2,1,34), (3,1,11).

Answer 2

One solution is easy:

Cut three 161s first:  5083cm-483cm=4600

Cut ten more: 4600-1610 = 2990

Cut 20 7s: 2990-140 = 2850

Cut 95 23s and 95 7s:  = 0

Answer 3

this is simple

please do your own homework