LCM

zooloo45 · #1 $Old$ May 26th, 2009, 03:38 AM

Find the number in which

14(n1)=33(n2)/2=23(n3)/2

where n1, n2 and n3 are positive whole number integers. If you can solve this algebraically, or logically without guess and check, i will be eternally grateful! Thank you!

Soroban · #2 $Old$ May 26th, 2009, 04:48 AM

Hello, zooloo45!

The problem is badly worded.

Quote:

Find the smallest positive integers $a,b,c$ so that: . $14a \:=\:\frac{33b}{2} \:=\:\frac{23c}{2}$

Multiply by 2: . $28a \;=\;33b \;=\;23c$

That is, we have an integer $N$ such that: . $N \;=\;2^2\!\cdot\!7\!\cdot\!a \;=\;3\!\cdot\!11\!\cdot\!b \;=\;23\!\cdot\!c$

The least $N$ occurs when: . $N \;=\;2^2\!\cdot\!3\!\cdot\!7\!\cdot\!11\!\cdot\!23 \;=\;21,\!252$

Therefore, we have: . $\begin{Bmatrix}28a \:=\:21,\!252 & \Longrightarrow & a \:=\:759 \\ 33b \:=\:21,\!252 & \Longrightarrow & b \:=\:644 \\ 23c \:=\:21,\!252 & \Longrightarrow & c \:=\:924 \end{Bmatrix}$