A number (≥3 digits) is divisible by 6 if its last two digits form a number divisible by 6.
Step 1: Know the rule for 6.
Divisibility by 6 means the number is divisible by both 2 and 3.
In math: \(6 = 2 \times 3\).
Step 2: What the statement claims.
It says: “Just check the last two digits.” If those two digits make a number divisible by 6, then the whole number is divisible by 6.
Step 3: Show a counterexample.
Take \(112\).
Last two digits = \(12\).
\(12 \div 6 = 2\), so \(12\) is divisible by 6.
Step 4: Check the real rule (2 and 3) on 112.
• Divisible by 2? Yes, because \(112\) is even.
• Divisible by 3? Check sum of digits: \(1 + 1 + 2 = 4\).
\(4\) is not a multiple of \(3\). So \(112\) is not divisible by \(3\).
Step 5: Conclude.
Since \(112\) is not divisible by 3, it is not divisible by 6 (even though its last two digits, \(12\), are divisible by 6).
Therefore, the statement is false.