How many positive integers between 1000 and 9999 inclusivea.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:Step-by-step explanation:a.first number is  1000-1+9=10089)1000(1     9-------      10        9     -----        10          9        ----          1        ----last number is 99999| 9999   ---------     1111 |0     --------9999=1008+(n-1)99999-1008=(n-1)9n-1=8991/9=999n=999+1=1000b.first digit=1000last digit=9999-1=99982 |9999   ---------   |4999|19998=1000+(n-1)2(n-1)2=9998-1000=8998n-1=4499n=4499=1=5000c.not sured.total  numbers=90009999=1000+(n-1)19999-1000=n-1n=8999+1=9000numbers divisible by 3=3000first number=1002last number=99999999=1002+(n-1)3(n-1)3=9999-1002=8997n-1=2999n=2999+1=3000numbers not divisible by 3=9000-3000=6000e. numbers divisible by 5=1800first number=1000last number=99959995=1000+(n-1)5(n-1)5=9995-1000=8995n-1=1799n=1799+1=1800numbers divisible by 7=12867 | 1000   ---------   |  142-61000-6+7=10017 | 9999   |---------     1428-39999-3=9996first digit=1001last digit=99969996=1001+(n-1)7(n-1)7=9996-1001=8995n-1=1285n=1285+1=1286numbers divisible by 35=257first digit=101535 ) 1000 ( 28         70        ----         300         280         ------            20            ---1000-20+35=101535)9999(285      70     ----      299      280      -----         199         175         ----           24          ----last digit=9999-24=99759975=1015+(n-1)35(n-1)35=9975-1015=8960n-1=8960/35=256n=257reqd. numbers=1800+1286-257=3019