Kenneth Rosen Edition 7 Exercise 6.1 Question 22 (Page No. 396)

Question

How many positive integers less than $1000$

• A. are divisible by $7?$
• B. are divisible by $7$ but not by $11?$ 
• C. are divisible by both $7$ and $11?$ 
• D. are divisible by either $7$ or $11?$ 
• E. are divisible by exactly one of $7$ and $11?$ 
• F. are divisible by neither $7$ nor $11? $
• G. have distinct digits?
• H. have distinct digits and are even?

Comments

Do you mean positive integers?,  because your question implies all integers, even negative, that would mean infinity for every part

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yes positive only

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https://gateoverflow.in/338113/kenneth-rosen-edition-7th-exercise-6-question-22-page-no-396

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https://gateoverflow.in/72351/counting

Answer 1

Rule : if numbers are in [1,n] then, total no of numbers divisible by k = $\left \lfloor \frac{n}{k} \right \rfloor$

1. $\left \lfloor \frac{999}{7} \right \rfloor$ = 142

2. divisible by 7 but not divisible by 11 => divisible by 7 but not divisible by 77

$\left \lfloor \frac{999}{7} \right \rfloor - \left \lfloor \frac{999}{77} \right \rfloor = 142 - 12 = 130$

3.$\left \lfloor \frac{999}{77} \right \rfloor = 12$

4. $\left \lfloor \frac{999}{7} \right \rfloor + \left \lfloor \frac{999}{11} \right \rfloor - \left \lfloor \frac{999}{77} \right \rfloor = 142 + 90 - 12 = 220$

5. $\left \lfloor \frac{999}{7} \right \rfloor + \left \lfloor \frac{999}{11} \right \rfloor - 2* \left \lfloor \frac{999}{77} \right \rfloor = 142 + 90 - 2*12 = 208$

[  $n(A\bigoplus B) = n(A)+n(B)-2n(A\cap B)$ ]

6. Total nos - numbers div either by 7 or 11 = 999 - 220 = 779

Comments

Thanks :D rest 2 ?

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last two needs, little bit of counting ! I will upload it later ! :) :)

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last 2 answers are:
g) positive integers less than 1000 having distinct digit.
it can be split into 3 cases:
# 3 digit number--9*9*8=648 integer (first digit can be chosen in 9 ways except 0 and second digit in 9 ways including zero and third digit in 8 ways)
#2 digit--9*9=81 integer (first digit can be chosen in 9 ways except 0 and second digit in 9 ways including zero)
#1 digit---9 integer (all integer from 1-9)

total integers=648+81+9=738 integer

h) positive integers less than 1000 having distinct digit and are even
# 3 digit no--9*8*1 ( 0 is chosen at unit place in 1 way and hundred place can be chosen in 9 ways and tens place in 8 ways)  + 8*8*4 ( 0 is not chosen at unit place(i.e 2,4,6,8 are chosen) in 4 way and hundred place can be chosen in 8 ways excluding 0 and tens place in 8 ways)
=9*8*1 + 8*8*4
=328 integer
# 2 digit no-- 9*1 ( 0 is chosen at unit place in 1 way and  tens place in 9 ways)  + 8*4 ( 0 is not chosen at unit place(i.e 2,4,6,8 are chosen) in 4 way and tens place in 8 ways excluding zero)
=9*1 + 8*4
=41 integer
# 1 digit no-4 integer(i.e 2,4,6,8 are chosen)

total integers=328+41+4=373

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you can make it answer !

Answer 2

Please Refer the following pdf

http://zimmer.csufresno.edu/~mnogin/math114fall04/41-18-sol.pdf

Answer 3

(A) In this case first integer is 7 and last integer is 994 and as per Arithmetic Progression we can write 7+(n-1)*7=994 which gives n=142 which is the answer.

(B) Ingers which are divisible by 7 i.e these 142 nos inger are to be sorted out which are not divisible by 11. we can do that by taking all multiples of LCM (7,11) i.e 77.

77 *1=77,77*2=154,77*3=231, 77*4=308, 77*5=385, 77*6=462,77*7=539,77*8=616,77*9=693,77*10=710,77*11=847,77*12=924

these 12 integers will excluded from 142 nos integers i.e 143-12=131 nos integers.

(C) integers  divisible both by 7 and 11  are the 12 nos as in (B) which are the multiples of LCM (7,11)

(D)Integers divisible  by  either 7 or 11.

Already we have got ingers divisible by 7 are 142 nos.

similarly we have to find out nos of integers divisible by 11 in which case first integer is 11 and last integer is 990. By A.P we can say

11+(n-1)*11=990 which gives n=90

so nos of integers divisible either by 7 or 11 is 142 +90=232

(E) Divisible exactly one of 7 and 11..

we know nos of ingers divisible by 7 is 142 nos

integers divisible by 11 is 90 nos

integers divisible both by 7 ans 11 is 12 nos,. hence nos of integers divisible one of 7 and 11 will 142+90-12=232-12=220 nos

(f) divisible neither by 7 nor 11

total nos of integers is 999

nos of integers divisible by 7 or 11 is 232.

so nos of integers divisible neither by 7 nor 11 is 999-232=767

(g) for distinct digit for two digit integers, total nos will be 10*9 =90 and for three digit integers it will be 10*9*8=720 and the total nos of integers will be =90+720=810 nos

(h) for distinct digits and even nos...in this case it will be 10*9

Answer 4

Points to be notice in these parts :

(D)are divisible by either 77 or 11? -→ Here, inclusive “OR” 

“Inclusive OR” means either 77 or 11 or both(7&11)

(E)are divisible by exactly one of 77 and 11? → Here, exclusive “OR” 

“Exclusive OR” means either 77 or 11 but not both(7&11) 

For detailed solutions of every parts refer the attachment provided below