Question

Find the smallest and largest four-digit numbers which when lessened by 12 are exactly divisible by \[16, 24, 40\].
\[
  (A)1208, 9848 \\
  (B)1200, 9840 \\
  (C)1212, 9852 \\
  (D)1188, 9828 \\
\]

Answer 1

Hint:
1) Least common multiple or smallest common multiple (denoted as L.C.M) of two integers (p and q) is the smallest integer that is divisible by both the integers p and q.
2) We can easily find the L.C.M of given integers simply by using the prime factorization method.

Complete step by step solution:
L.C.M of 16\[ = 2 \times 2 \times 2 \times 2\]
L.C.M of 24 \[ = 2 \times 2 \times 2 \times 3\]
L.C.M of 40 \[ = 2 \times 2 \times 2 \times 5\]
L.C.M of 16, 24 and 40\[ = 2 \times 2 \times 2 \times 2 \times 3 \times 5 = 240\].
240 is the smallest number which will completely divide the numbers (16, 24 and 40).
But as 240 is not a four-digit number.
\[ \Rightarrow \] Multiples of 240 are also completely divisible by the numbers (16, 24 and 40).
Four-digits multiples of 240 are \[1200, 1440, 1680..., 9600, 9840\].
Smallest number is \[1200\] that completely divides the numbers (16, 24 and 40).
So, our required number is \[1200 + 12 = 1212\].
Greatest number is \[9840\] that completely divides the numbers (16, 24 and 40).
So, our required number is \[9840 + 12 = 9852.\]

Option (C) is correct.

Note:
The below mentioned points will help you to solve this kind of questions:
1) Whenever you are required to find a number that is divisible by more than one number, find LCM.
2) Whenever you are required to find a number that completely divides more than one number, find HCF.