Answer
a) 142
b) 130
c) 12
d) 220
e) 208
f) 779
g)738
h) 373
Work Step by Step
a) Integers divisible by 7 that are less than 1000 are number of integers that are multiple of 7 and are less than 1000.
Therefore, such numbers are $\left \lfloor{\frac{999}{7}}\right \rfloor$ =142.
We have taken 999 because we want integers less than 1000.
b) We want to find integers that are divisible by 7 but not by 11 i.e., we can say that not divisible by 77 or integers that are multiple of 7 but not of 11.
Therefore, such numbers are $\left \lfloor{\frac{999}{7}}\right \rfloor$-$\left \lfloor{\frac{999}{77}}\right \rfloor$ =142-12=130.
c) Integers that are both divisible by both 7 and 11 are those integers that are divisible by lcm of 7 and 11 i.e., divisible by 77.
Therefore, such numbers are $\left \lfloor{\frac{999}{77}}\right \rfloor$=12.
d) We want to find the integers that are divisible by 7 and 11.
By principle of inclusion-exclusion, Numbers of such integers is given by
Number of integers divisible by 7 + Number of integers divisible by 11- Number of integers divisible by 77
Ans=$\left \lfloor{\frac{999}{7}}\right \rfloor$+$\left \lfloor{\frac{999}{11}}\right \rfloor$-$\left \lfloor{\frac{999}{77}}\right \rfloor$=142+90-12=220
e) First, number of integers that are divisible by 7 but not by 11 =130
Secondly, number of integers that are divisible by 11 but not by 7 are integers that are divisible by 11 but not by 77.
As we did in part b, number of such integers = $\left \lfloor{\frac{999}{11}}\right \rfloor$-$\left \lfloor{\frac{999}{77}}\right \rfloor$=90-12=78.
Therefore, answer =130+78=208.
f) Integers divisible by neither 7 nor 11 are the integers in the set that are not divisible either 7 or 11.
Therefore such integers= total number of integers- integers divisible by either 7 or 11
=999-220-779.
g) Integers less than 1000 can have 3, 2 or 1 digit.
Case 1- 1 digit
$\quad$ There are 9 possible integers less than 1000 that have 1 digit $\quad$:1,2,3,4,5,6,7,8,9
Case 2- 2 digits
$\quad$Tens place: This place cannot be 0. So, we have 9 possible values $\quad$(1,2,3,4,5,6,7,8,9)
$\quad$Ones place: Since this digit cannot be same as the tens place, so again we$\quad$ have 9 possible choices for this digit.
$\quad$Use the product rule 9.9 =81 ways.
Case 3- 3 digits
$\quad$ Hundreds place: We have 9 choices because this digit cannot be 0.
$\quad$ Tens place: We have 9 choices because this digit cannot be same as the
$\quad$ hundreds place.
$\quad$ Ones place- Since this integer cannot be same as the hundreds or tens $\quad$digit, so we have 8 options for this digit.
$\quad$Use the product rule: 8.9.9=648
$\quad$In total- 9+81+648= 738
h) Integers below 1000 have 3 or less digits.
Case 1- 1 digit
$\quad$ There are 4 possible even integers: 2,4,6,8
Case 2- 2 digit
$\quad$ There are again 2 possible.
$\quad$- If tens digit is odd. We have 5 possible choices for tens digit and last digit$\quad$ has to be one of 0,2,4,6,8.
$\quad$- If tens digit is even. We have 4 possible choices for tens digit (2,4,6,8) and $\quad$last digit can be any even number other than the tens digit because we $\quad$want distinct digit.
$\quad$ Now use the product rule and sum rule=5.5+4.4=25+16=41 ways.
Case 3- 3 digit
-Hundreds place- 9 choices out of which 5 are odd and 4 are even.
- Tens place- 9 ways since the tens place cannot be same as hundreds place.
If the hundreds place is odd :4 odd and 5 even.
If the hundreds place is even :5 odd and 4 even.
-Ones place
If the first two digits are odd: we have 5 choices since this digit need to be even.
If the first two digits are even: we have 3 choices
If the first digit is odd and second is even: 4 ways
If the first digit is even and second is odd: 4 ways
Now use the product rule:
- first two digits odd: 5.4.5=100
- first two digits even : 4.4.3= 48
- first digit even and second digit odd- 4.5.4=80 ways
- first digit odd and second digit even- 5.5.4=100 ways
Use sum rule =100+100+48+80=328
In total, 4+41+328=373 ways.
