Question

How many positive integers between 1000 and 9999 inclusive

a.are divisible by 9?


b.are even?


c.have distinct digits?


d.are not divisible by 3?


e.are divisible by 5 or 7? f) are divisible by 5 but not by 7? g) are divisible by 5 and 7?

Answer 1

Explanation:

a.

first number is 1000-1+9=1008

9)1000(1

9

-------

10

9

-----

10

9

----

1

----

last number is 9999

9| 9999

---------

1111 |0

--------

9999=1008+(n-1)9

9999-1008=(n-1)9

n-1=8991/9=999

n=999+1=1000

b.

first digit=1000

last digit=9999-1=9998

2 |9999

---------

|4999|1

9998=1000+(n-1)2

(n-1)2=9998-1000=8998

n-1=4499

n=4499=1=5000

c.not sure

d.

total numbers=9000

9999=1000+(n-1)1

9999-1000=n-1

n=8999+1=9000

numbers divisible by 3=3000

first number=1002

last number=9999

9999=1002+(n-1)3

(n-1)3=9999-1002=8997

n-1=2999

n=2999+1=3000

numbers not divisible by 3=9000-3000=6000

e.

numbers divisible by 5=1800

first number=1000

last number=9995

9995=1000+(n-1)5

(n-1)5=9995-1000=8995

n-1=1799

n=1799+1=1800

numbers divisible by 7=1286

7 | 1000

---------

| 142-6

1000-6+7=1001

7 | 9999

|---------

1428-3

9999-3=9996

first digit=1001

last digit=9996

9996=1001+(n-1)7

(n-1)7=9996-1001=8995

n-1=1285

n=1285+1=1286

numbers divisible by 35=257

first digit=1015

35 ) 1000 ( 28

70

----

300

280

------

20

---

1000-20+35=1015

35)9999(285

70

----

299

280

-----

199

175

----

24

----

last digit=9999-24=9975

9975=1015+(n-1)35

(n-1)35=9975-1015=8960

n-1=8960/35=256

n=257

reqd. numbers=1800+1286-257=3019