Question:

Let $5000<N<9000$ and $N$ has digits from the set $\{0,1,2,5,9\}$. If digits can be repeated, then find the number of such numbers $N$ which are divisible by $3$.

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For divisibility problems involving digit sets, always classify digits by their remainders modulo the required divisor.
Updated On: Jan 25, 2026
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Correct Answer: 84

Solution and Explanation

Step 1: Identify the form of $N$.
Since $5000 Step 2: Use the divisibility rule of $3$. 
A number is divisible by $3$ if and only if the sum of its digits is divisible by $3$. 
Digits available: \[ \{0,1,2,5,9\} \] Residues modulo $3$: \[ 0:\{0,9\},\quad 1:\{1\},\quad 2:\{2,5\} \] Step 3: Count valid combinations of the last three digits. 
For each fixed thousands digit, count the number of ordered triples $(a,b,c)$ such that: \[ \text{(digit sum)} \equiv 0 \pmod{3} \] Total valid combinations for the last three digits: \[ 42 \] Step 4: Multiply by the choices for the thousands digit. 
\[ \text{Total numbers}=2\times 42=84 \]

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