# For all possible numbers using the digits 0,1,3,4,7 (a)how many numbers are greater than 4000? (b)how many numbers have 3 as the first digit and 0 as the last? (c)how many are divisible by four?

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### 3 Answers

For all possible numbers using the digits 0,1,3,4,7

(a)how many numbers are greater than 4000?

All possible numbers with digits 0, 1 , 3 , 4 and 7 greater are equals to 4000= 2 x 5 x 5 x 5 =250

(first digit can be either 4 or 7 , remaing digits can be either 0, 1,3 ,4 or 7)

**Number greater than 4000 = 250-1 =249**

because one number formed=4000 ,which is equal to 4000.

(b)how many numbers have 3 as the first digit and 0 as the last?

Number with 3 as the first digit and 0 as last =1 x 5 x 5 x 1

=25

(first digit can be 3 , in between two digits can be either 0, 1,3 ,4 or 7 and last digit only 0 )

(c)how many are divisible by four?

If last two digits number is divisible by 4 then number will be divisible by 4/

Last two digits can be formed which is divisible by 4 are 00,04,40,44

= 2 x 2

=4

Thus numbers divisible by 4 are= 5 x 5 x 2 x 2 = 100

Here 0 is also cosidered at first digit.

If 0 is not considered as first digit then numbers are

=4 x 5 x 2 x2=80

(a)Using 0,1,3,4,7-- numbers greater than 4000(without repeating the numbers)

First for four digit numbers:

Frist digit canbe 4 or 7.

possibilities for first digit=2

possibilities for second digit=4

possibilities for 3rd digit=3

possibilities for fourth digit=2

Hence possibilities for numbers greater than 4000=2x4x3x2=48

Secondly For five digit numbers

Frist digit canbe 1, 3,4 or 7, possibilities for first digit=4

possibilities for second digit=4

possibilities for 3rd digit=3

possibilities for fourth digit=2

possibilities for fifth digit=1

Hence possibilities for five digit numbers greater than 4000=4x4x3x2x1=96

total numbers possible=48+96=144

(b)

numbers with 3 as first digit and 0 as the last digità

such 2 digit numbers can be: 1

such 3 digit numbers without repetition can be : 3

such 4 digit numbers can be:3__ __0: 6

such 5 digit numbers can be: 3__ __ __ 0: 6

total 1+3+6+6=16

(c)to be divisible by 4, last 2 digits have to be divisible by 4

various numbers can be

4

40

44

3 digit numbers can be _40: 3

_44: 3

4 digit numbers can be __ __ 40: 3x2=6

__ __ 44 : 3x2=6

5 digit numbers can be __ __ __ 40: 3x2x1=6

__ __ __ 44: 3x2x1= 6

total=1+1+1+3+3+6+6+6+6=33

This is the correct answer

(a)Using 0,1,3,4,7-- numbers greater than 4000(without repeating the numbers)

First for four digit numbers:

Frist digit canbe 4 or 7.

possibilities for first digit=2

possibilities for second digit=4

possibilities for 3rd digit=3

possibilities for fourth digit=2

Hence possibilities for numbers greater than 4000=2x4x3x2=48

Secondly For five digit numbers

Frist digit canbe 1, 3,4 or 7, possibilities for first digit=4

possibilities for second digit=4

possibilities for 3rd digit=3

possibilities for fourth digit=2

possibilities for fifth digit=1

Hence possibilities for five digit numbers greater than 4000=4x4x3x2x1=96

total numbers possible=48+96=144