Question : If the number formed by the last two digits of a three-digit integer is an integral multiple of 6, the original integer itself will always be divisible by:
Option 1: 6
Option 2: 3
Option 3: 2
Option 4: 4
Correct Answer: 2
Solution :
Given: The last two digits of a three-digit integer is an integral multiple of 6.
Let the 3-digit integer be $100x+10y+z$.
$10y+z =6m$ ($m$ is any integer)
The 3-digit integer is $100x+10y+z = 100x+6m = 2(50x+3m)$.
So, the original integer itself will always be divisible by 2.
Hence, the correct answer is 2.
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