How many positive integers between 1000 and 9999 inclusive
       
      
                
     
    
    
    
    
    1 answer:
            
              
              
                
                
Answer:
Step-by-step 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
 
                                
             
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