To solve the problem of finding the least number such that when 2 is subtracted from it, the resulting difference is exactly divisible by 8 and 12, we can follow these steps:
1. Understand the Requirement:
- We need a number [tex]\( x \)[/tex] such that [tex]\( x - 2 \)[/tex] is divisible by both 8 and 12.
2. Common Multiple:
- The number [tex]\( x - 2 \)[/tex] must be a multiple of a common multiple of 8 and 12. Hence, we need to find the least common multiple (LCM) of 8 and 12.
3. Finding the LCM:
- The least common multiple of 8 and 12 is the smallest number that is a multiple of both 8 and 12.
- The multiples of 8 are 8, 16, 24, 32, etc.
- The multiples of 12 are 12, 24, 36, 48, etc.
- The smallest common multiple of 8 and 12 is 24.
4. Forming the Required Number:
- Let’s denote the LCM by [tex]\( \text{LCM} \)[/tex]. Here, the [tex]\(\text{LCM} = 24\)[/tex].
- We need to form the number [tex]\( x \)[/tex] such that [tex]\( x - 2 = \text{LCM} \)[/tex].
- Therefore,
[tex]\[
x - 2 = 24
\][/tex]
5. Solving for [tex]\( x \)[/tex]:
- Adding 2 to both sides of the equation, we get:
[tex]\[
x = 24 + 2
\][/tex]
[tex]\[
x = 26
\][/tex]
So, the least number [tex]\( x \)[/tex] from which when 2 is subtracted, the difference is exactly divisible by both 8 and 12, is:
[tex]\[
\boxed{26}
\][/tex]
