Asked by Sandy

Kenji has a box containing many blue rods, each one 252 cm long. Mei-Ling has a box containing many green rods, each one 408 cm long. They each start making a long line of rods of their colour by laying rods end to end. After a while, their lines are exactly the same length. What is this length, if Kenji and Mei-Ling have each used the minimum number of rods? How many rids did each use? Five a complete, well-organized solution.
Answered by Damon
you are looking for the least common multiple of 252 and 408

The way I do such problems is to factor each
252 = 2*126
= 2*2*63
= 2*2*3*21
= 2*2*3*3*7
then
408 = 2*204
= 2*2*102
= 2*2*2*51

so we need
2*2*2 * 3*3 * 7 * 51 = 8*9*7*51=25704
check
25704 / 252 = 102 blue
25704 / 408 = 63 green
Answered by Lola
Isn't 51 divisible by 3 and 17?
Answered by Roblox
Yea factor each 102 blue, 63 green, also please try my game out "roblox", thanks!
Answered by Lucas
you didnt make roblox, people always try to take credit these days,
Answered by Remy
idk
Answered by kjhg6yn
mmmmmmmmmmmmmmmmmmmmm
Answered by Ali
with correction (question says minimum rods)

252 = 2*126
= 2*2*63
= 2*2*3*21
= 2*2*3*3*7
then
408 = 2*204
= 2*2*102
= 2*2*2*51
= 2*2*2*17*3

so we need
2*2*2 * 3*3 * 7 * 17 = 8*9*7*17 = 8568
check
8568 / 252 = 34 blue
8568 / 408 = 21 green
Answered by Bot
Is there anything else I can help you with?
Answered by Bot
You are absolutely correct! Thank you for pointing out my mistake. The least common multiple of 252 and 408 is indeed 8568, and the minimum number of rods needed by Kenji and Mei-Ling are 34 and 21 respectively.