Question

Find the smallest number of six digits divisible by 18, 24 and 30.

Answer 1

Hint: Let l be the LCM (a,b,c). Then $\forall n$ such that $\left. a \right|n,\left. b \right|n$ and $\left. c \right|n\Rightarrow \left. l \right|n$ . Hence first find the LCM of 18,24 and 30. Then find the smallest six digit multiple of the LCM and hence the number obtained will be the smallest six-digit number divisible by 18, 24 and 20.

Complete step-by-step solution -
we know that
Let l be the LCM (a,b,c). Then $\forall n$ such that $\left. a \right|n,\left. b \right|n$ and $\left. c \right|n\Rightarrow \left. l \right|n$ . Hence the smallest six-digit number divisible by 18,24 and 30 will be the smallest six digit multiple of LCM(18,24,30).
$\begin{align}
  & 18=2\times {{3}^{2}} \\
 & 24={{2}^{3}}\times 3 \\
 & 20={{2}^{2}}\times 5 \\
\end{align}$
LCM(18,20,24) $={{2}^{3}}\times {{3}^{2}}\times 5=360$
Method for finding smallest n-digit number divisible by k:
Step I: Write the smallest n digit number. Let that number = p
Step II: Divide p by k to get remainder r
Step III: The smallest n-digit number divisible by k = p+(k-r)
Smallest 6 digit number = 1,00,000
So, we have p = 1,00,000 and k = 360
We know$100000=360\times 277+280$
Hence r = 280
Hence the smallest 6 digit number divisible by 360 = 1,00,000+(360-280) = 1,00,000+80=1,00,080
Hence the smallest 6-digit number divisible by 18,24,30 is 1,00,080.

Note: Let l = lcm(a,b) and let n be an integer such that $\left. a \right|n$ and $\left. b \right|n$ .
Our claim is that l divides n.
Proof:
Since l is the LCM of a and b $\forall m\in \mathbb{N}$ such that $\left. a \right|m$ and $\left. b \right|m\Rightarrow l\le m$ , we have
$n\ge l$ .
Let us assume that l does not divide n.
Using Euclid's division lemma we have
$\begin{align}
  & n=ql+r,0 < r < l \\
 & \Rightarrow n-ql=r \\
\end{align}$
Since $a|n$ and$a|l$, we have $a|n-ql$and hence$a|r$.