How many 3-digit numbers are completely divisible by 6?

### Concept & Formula To find the count of numbers in an Arithmetic Progression (A.P.) that are divisible by a certain number, we first identify the smallest and largest numbers in that range divisible by the target number. The number of terms $n$ in an A.P. is calculated using the formula: $$n = \frac{L - A}{D} + 1$$ Where: * $L$ = Last term * $A$ = First term * $D$ = Common difference ### Step-by-Step Solution * **Given:** The range is all 3-digit numbers, which spans from $100$ to $999$. The divisor is $6$. * **Step 1: Find the first 3-digit number divisible by 6.** Divide $100$ by $6$. The remainder is $4$. To make it divisible by $6$, we must add $(6 - 4) = 2$. So, the first term $A = 100 + 2 = 102$. * **Step 2: Find the last 3-digit number divisible by 6.** Divide $999$ by $6$. The remainder is $3$. Subtract the remainder from $999$ to get the largest multiple. So, the last term $L = 999 - 3 = 996$. * **Step 3: Calculate the total number of terms.** Using the A.P. formula with $D = 6$: $$n = \frac{996 - 102}{6} + 1$$ $$n = \frac{894}{6} + 1$$ $$n = 149 + 1 = 150$$ ### Exam Strategy & Shortcut For any range starting from $1$ to $N$, the number of multiples of $X$ is simply the integer part of $\frac{N}{X}$. To find multiples specifically for 3-digit numbers, subtract the multiples up to $99$ from the multiples up to $999$: Multiples up to $999 = \lfloor \frac{999}{6} \rfloor = 166$ Multiples up to $99 = \lfloor \frac{99}{6} \rfloor = 16$ Total = $166 - 16 = 150$. This is much faster than finding the exact boundary terms! ### Common Pitfall Students often forget to add $1$ at the end of the A.P. formula, calculating $\frac{L - A}{D}$ and stopping there, which leads to selecting $149$ instead of $150$. Always remember that inclusive boundary counting requires the $+1$. ### Final Answer **Therefore, the correct answer is 150.**