Question

How many four-digit numbers are divisible by 5, 12, and 18.
(a) 48
(b) 49
(c) 47
(d) 50

Answer 1

Hint: To solve this question we will use the concept of LCM and AP. LCM is short of the least common multiple which is the smallest possible number which is divisible by all given numbers.
AP is short for arithmetic progression whose \[{{n}^{th}}\] term is determined by the formula, \[{{a}_{n}}={{a}_{1}}+\left( n-1 \right)d\], where \[{{a}_{n}}={{n}^{th}}\]term, \[{{a}_{1}}\to \] first number, \[n\to \] number of terms & d is common difference.

Complete step-by-step solution:
We have to find all possible four-digit numbers divisible by 5, 12, and 18.
We first factor 5 and 12 and 18.
Factors of 5 are,
\[5=1\times 5\]
Factors of 12 are,
\[12=2\times 2\times 3={{2}^{2}}\times 3\]
Factors of 18 are,
\[18=2\times 3\times 3={{3}^{2}}\times 2\]
Now we will proceed to find LCM (Least common multiple) of 5, 12 and 18.
LCM of these are given as,
\[\begin{align}
  & 2\left| \!{\underline {\,
  5,12,18 \,}} \right. \\
 & 3\left| \!{\underline {\,
  5,6,9 \,}} \right. \\
 & 3\left| \!{\underline {\,
  5,2,3 \,}} \right. \\
 & 2\left| \!{\underline {\,
  5,2,1 \,}} \right. \\
 & 5\left| \!{\underline {\,
  5,1,1 \,}} \right. \\
 & \left| \!{\underline {\,
  1,1,1 \,}} \right. \\
\end{align}\]
So, LCM of 5, 12 and 18 = \[2\times 2\times 3\times 3\times 5\]
LCM of 5, 12 and 18 = \[{{2}^{2}}\times {{3}^{2}}\times 5=180\]
Because our 4 digit number must be divisible by 5, 12 and 18 therefore it should be divisible by their LCM which is 180.
Hence our 4 digit number must be divisible by 180.
Now the smallest 4 digit number is 1000.
We need a number to be divisible by 180. Hence, we will divide 1000 by 180.
\[\Rightarrow \dfrac{1000}{180}=5.55\cong 6\]
Therefore the smallest 4 digit number divisible by 180 can be obtained by multiplying 6 to 180.
\[\Rightarrow 180\times 6=1080\] ---------- (1)
So, 1080 is the smallest four-digit number divisible by 180.
Similarly, the largest 4 digit number is 9999.
Dividing 9999 by 180 gives,
\[\dfrac{9999}{180}\cong 55\]
\[\therefore \] Highest 4 digit number that is divisible by 180 is,
\[180\times 55=9900\] -------------- (2)
Using the terms obtained in equation (1) and (2). We can form an AP (Arithmetic progression) of terms that are divisible by 180.
Then the AP is,
1080, 1260, …… 9900
Here first term, \[{{a}_{1}}=1080\].
Common difference, d = 1260 – 1080 = 180.
And \[{{n}^{th}}\] term, \[{{a}_{n}}\] = 9900.
We have to find the value of n.
Formula of AP \[{{n}^{th}}\] term is given as,
\[{{a}_{n}}={{a}_{1}}+\left( n-1 \right)d\]
Substituting all values we get;
9900 = 1080 + (n - 1)180
Subtracting 1080 both sides we get,
$(n - 1)180 = 9900 – 1080$
\[\Rightarrow (n - 1)180 = 8820 \]
Dividing by 180 we get,
\[\begin{align}
  & \left( n-1 \right)=\dfrac{8820}{180} \\
 & \Rightarrow \left( n-1 \right)=49 \\
\end{align}\]
Adding 1 on both sides,
\[\Rightarrow \] $n = 49 + 1 = 50$
Therefore, value of $n = 50.$
Hence the total number of 4 digit numbers which are divisible by 5, 12, and 18 are 50, which is an option (d).

Note: Approximating the largest and smallest possible four-digit number divisible by 180 is correct because any two consecutive numbers divisible by 180 have at least a difference of 180 in between. Therefore approximation works well in this case. Also, any number divisible by 180 (LCM of 5, 12, and 18) is also divisible by one or all of 5, 12, and 18. So this method works well.