Maths Q14 – AP-EAMCET 2026

Home Maths Permutations and CombinationsQuestion #311

If all possible 6 digit numbers are formed by using all the digits 1, 3, 5, 6, 7, 9 without repeating any digit, then the number of numbers which are greater than 3,00,000 and divisible by 4 is

Clue 1 of 3: Divisibility Rule for 4

A number is divisible by 4 if the number formed by its last two digits is divisible by 4.

🧠 Full Solution Path

Step 1: Identify Digits and Constraints✦ Active

The given distinct digits are $D = {1, 3, 5, 6, 7, 9}$ . We need to form 6-digit numbers without repetition. The two conditions are:

1. The number must be greater than 3,00,000. This implies the first digit ( $d_{1}$ ) cannot be 1. So, $d_{1} \in {3, 5, 6, 7, 9}$ .

2. The number must be divisible by 4. This implies the number formed by its last two digits ( $d_{5} d_{6}$ ) must be divisible by 4.

💡 Teacher's Secret Hint

Remember that for a 6-digit number $d_{1} d_{2} d_{3} d_{4} d_{5} d_{6}$ , the 'greater than 3,00,000' condition only depends on the first digit $d_{1}$ and its relation to 3.

Step 2: Determine Possible Last Two Digits (

d_{5} d_{6}

)○ Expand

We list all two-digit numbers formed from $D = {1, 3, 5, 6, 7, 9}$ that are divisible by 4:

- Numbers ending in 1: None (e.g., $X 1$ , no multiple of 4)

- Numbers ending in 3: None (e.g., $X 3$ , no multiple of 4)

- Numbers ending in 5: None (e.g., $X 5$ , no multiple of 4)

- Numbers ending in 6: $16, 36, 56, 76$ (all are divisible by 4 and use digits from $D$ )

- Numbers ending in 7: None (e.g., $X 7$ , no multiple of 4)

- Numbers ending in 9: None (e.g., $X 9$ , no multiple of 4)

So, the possible last two digits ( $d_{5} d_{6}$ ) are $16, 36, 56, 76$ . We will analyze each of these 4 cases.

💡 Teacher's Secret Hint

Carefully check all combinations. For example, $92$ is divisible by 4, but digit '2' is not in the given set.

Step 3: Analyze Cases for

d_{1}

and Remaining Digits○ Expand

Let the 6-digit number be $d_{1} d_{2} d_{3} d_{4} d_{5} d_{6}$ . For each $d_{5} d_{6}$ pair, we determine the available digits for $d_{1}$ (must be $\geq 3$ ) and then arrange the remaining 3 digits for $d_{2} d_{3} d_{4}$ .

Case 1: $d_{5} d_{6} = 16$ . Digits used: ${1, 6}$ . Remaining digits for $d_{1} d_{2} d_{3} d_{4}$ : ${3, 5, 7, 9}$ .

For $d_{1}$ , valid choices are ${3, 5, 7, 9}$ (since all are $\geq 3$ ). So, 4 choices for $d_{1}$ .

The remaining 3 digits for $d_{2} d_{3} d_{4}$ can be arranged in $3! = 6$ ways.

Number of numbers in this case = $4 \times 3! = 4 \times 6 = 24$ .

Case 2: $d_{5} d_{6} = 36$ . Digits used: ${3, 6}$ . Remaining digits for $d_{1} d_{2} d_{3} d_{4}$ : ${1, 5, 7, 9}$ .

For $d_{1}$ , valid choices are ${5, 7, 9}$ (digit 1 is excluded as $1 X X X X X < 300000$ ). So, 3 choices for $d_{1}$ .

The remaining 3 digits for $d_{2} d_{3} d_{4}$ can be arranged in $3! = 6$ ways.

Number of numbers in this case = $3 \times 3! = 3 \times 6 = 18$ .

Case 3: $d_{5} d_{6} = 56$ . Digits used: ${5, 6}$ . Remaining digits for $d_{1} d_{2} d_{3} d_{4}$ : ${1, 3, 7, 9}$ .

For $d_{1}$ , valid choices are ${3, 7, 9}$ (digit 1 is excluded). So, 3 choices for $d_{1}$ .

The remaining 3 digits for $d_{2} d_{3} d_{4}$ can be arranged in $3! = 6$ ways.

Number of numbers in this case = $3 \times 3! = 3 \times 6 = 18$ .

Case 4: $d_{5} d_{6} = 76$ . Digits used: ${7, 6}$ . Remaining digits for $d_{1} d_{2} d_{3} d_{4}$ : ${1, 3, 5, 9}$ .

For $d_{1}$ , valid choices are ${3, 5, 9}$ (digit 1 is excluded). So, 3 choices for $d_{1}$ .

The remaining 3 digits for $d_{2} d_{3} d_{4}$ can be arranged in $3! = 6$ ways.

Number of numbers in this case = $3 \times 3! = 3 \times 6 = 18$ .

💡 Teacher's Secret Hint

Pay close attention to which digits are *remaining* after $d_{5} d_{6}$ are chosen, as these will affect the available choices for $d_{1}$ and the arrangements for $d_{2} d_{3} d_{4}$ .

Step 4: Calculate Total Number of Numbers○ Expand

Summing the numbers from all valid cases:

T o t a l = 24 (f r o m C a s e 1) + 18 (f r o m C a s e 2) + 18 (f r o m C a s e 3) + 18 (f r o m C a s e 4)

T o t a l = 24 + 54 = 78

Now, we check which option matches this value:

1. $5 \times 4! = 5 \times 24 = 120$

2. $5! \times 4! = 120 \times 24 = 2880$

3. $13 \times 3! = 13 \times 6 = 78$

4. $4 \times 4! = 4 \times 24 = 96$

The total number of such numbers is 78, which corresponds to option 3.

💡 Teacher's Secret Hint

Ensure to verify your final calculated sum against the given options, often the options are presented in factored form ( $n \times k!$ ).

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Maths Question 14 – AP-EAMCET 2026

Clue 1 of 3: Divisibility Rule for 4

🧠 Full Solution Path