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1

AIEEE 2002

Five digit number divisible by 3 is formed using 0, 1, 2, 3, 4 and 5 without repetition. Total number of such numbers are
A
312
B
3125
C
120
D
216

Explanation

Note : For a number to be divisible by 3, the sum of digits should be divisible by 3.

Here given numbers are 0, 1, 2, 3, 4 and 5. Out of those 6 numbers possible sets of 5 numbers are (1, 2, 3, 4, 5) and (0, 1, 2, 4, 5) whose sum are divisible by 3.

Set 1 : Set is = (1, 2, 3, 4, 5). Sum of digits = 1 + 2 + 3 + 4 + 5 = 15 (Divisible by 3)

So total no of arrangement = 1$$\times$$2$$\times$$3$$\times$$4$$\times$$5 = 5!

Set 2 : Set is = (0, 1, 2, 4, 5). Sum of digits = 0 + 1 + 2 + 4 + 5 = 12 (Divisible by 3)

So total no of arrangement = 4$$\times$$4$$\times$$3$$\times$$2$$\times$$1 = 4.4!

$$\therefore$$ Total arrangement = 5! + 4.4! = 216
2

AIEEE 2002

Number greater than 1000 but less than 4000 is formed using the digits 0, 1, 2, 3, 4 (repetition allowed). Their number is
A
125
B
105
C
374
D
625

Explanation

There are 3 possible ways that we can make number greater than 1000 but less than 4000 using the digits 0, 1, 2, 3, 4 where repetition is allowed

Case 1 : First digit is 1 = 1 _ _ _

Possible numbers starting with 1 = 1$$\times$$5$$\times$$5$$\times$$5 = 125

But this includes 1000 also which does not satisfy the given condition of being greater than 1000. Hence there will be 124 numbers having 1 in the first place.

Case 2 : First digit is 2 = 2 _ _ _

Possible numbers starting with 2 = 1$$\times$$5$$\times$$5$$\times$$5 = 125

Case 3 : First digit is 3 = 3 _ _ _

Possible numbers starting with 3 = 1$$\times$$5$$\times$$5$$\times$$5 = 125

Total possible numbers = 124 + 125 + 125 = 374
3

AIEEE 2002

Total number of four digit odd numbers that can be formed using 0, 1, 2, 3, 5, 7 (using repetition allowed) are
A
216
B
375
C
400
D
720

Explanation

$$\therefore$$ Total no of ways = 5$$\times$$6$$\times$$6$$\times$$$${}^4{C_1}$$ = 720

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