Find the number of integers lying between 100
and 400 (excluding the two) and divisible by any one of the numbers 2 or 3 or 5
or 7?

Solution follows here:

__Solution:__

__Method-I____(__Set theory method)

This method is a traditional mathematical
model. This is based on union of four sets. This is a laborious method not
suitable for competitive exams given the availability of limited time.

Follow the notation:

n(x) -> number of integers lying between
100 and 400 and divisible by x

n(x ∩ y) ->
number of integers lying between 100 and 400 and divisible by x and y

**Number of integers divisible by 2: n(2)**

102,104,……398 – all even integers starting
from 102 and ending with 398

This is an A.P with first term 102, last term
398 and common difference 2.

If n terms are considered in A.P, last_term = t

_{n}= a+(n-1)d = first_term + (n-1)d
=>

**n = (last_term - first_term)/d + 1**
n(2) = (398-102)/2 + 1 = 296/2 + 1 = 149

**Number of integers divisible by 3: n(3)**

102,105,…..399

n(3) = (399-102)/3 + 1 = 297/3 + 1 = 100

**Number of integers divisible by 5: n(5)**

105,110,…..395

n(5) = (395-105)/5 + 1 = 290/5 + 1 = 59

**Number of integers divisible by 7: n(7)**

105,110,…..399

n(7) = (399-105)/7 + 1 = 294/7 + 1 = 43

**Number of integers divisible by both 2 and 3: n(2**

**∩**

**3)**

As 2 and 3 have no common factors, n(2 ∩ 3) = n(6)

102,108,…..396

n(2 ∩ 3) =
(396-102)/6 + 1 = 294/6 + 1 = 50

**Number of integers divisible by both 2 and 5: n(2**

**∩**

**5)**

n(2 ∩ 5) = n(10)

110,120,…..390

n(2 ∩ 5) =
(390-110)/10 + 1 = 280/10 + 1 = 29

**Number of integers divisible by both 2 and 7: n(2**

**∩**

**7)**

n(2 ∩ 7) = n(14)

112,126,…..392

n(2 ∩ 7) =
(392-112)/14 + 1 = 280/14 + 1 = 21

**Number of integers divisible by both 3 and 5: n(3**

**∩**

**5)**

n(3 ∩ 5) = n(15)

105,120,…..390

n(3 ∩ 5) = (390-105)/15
+ 1 = 285/15 + 1 = 20

**Number of integers divisible by both 3 and 7: n(3**

**∩**

**7)**

n(3 ∩ 7) = n(21)

105,126,…..399

n(3 ∩ 7) =
(399-105)/21 + 1 = 294/21 + 1 = 15

**Number of integers divisible by both 5 and 7: n(5**

**∩**

**7)**

n(5 ∩ 7) = n(35)

105,140,…..385

n(5 ∩ 7) =
(385-105)/35 + 1 = 280/35 + 1 = 9

**Number of integers divisible by 2 and 3 and 5: n(2**

**∩**

**3**

**∩**

**5)**

n(2 ∩ 3 ∩ 5) = n(30)

120,150,…..390

n(2 ∩ 3 ∩ 5) = (390-120)/30 + 1 = 270/30 + 1 = 10

**Number of integers divisible by 2 and 3 and 7: n(2**

**∩**

**3**

**∩**

**7)**

n(2 ∩ 3 ∩ 7) = n(42)

3*42, 4*42, 5*42, 6*42,,….. 9*42

n(2 ∩ 3 ∩ 7) = 7

**Number of integers divisible by 2 and 5 and 7: n(2**

**∩**

**5**

**∩**

**7)**

n(2 ∩ 5 ∩ 7) = n(70)

2*70, 3*70, 4*70, 5*70

n(2 ∩ 5 ∩ 7) = 4

**Number of integers divisible by 3 and 5 and 7: n(3**

**∩**

**5**

**∩**

**7)**

n(3 ∩ 5 ∩ 7) = n(105)

1*105, 2*105, 3*105

n(3 ∩ 5 ∩ 7) = 3

**Number of integers divisible by 2 and 3 and 5 and 7: n(2**

**∩**

**3**

**∩**

**5**

**∩**

**7)**

n(2 ∩ 3 ∩ 5 ∩ 7) = n(210)

1*210 is the only number

n(2 ∩ 3 ∩ 5 ∩ 7) = 1

using the formula

**n(A****U****B****U****C****U****D)****= n(A) + n(B) + n(C) + n(D) – n(A**

**∩**

**B) – n(A**

**∩**

**C) – n(A**

**∩**

**D) – n(B**

**∩**

**C) – n(B**

**∩**

**D) – n(C**

**∩**

**D) + n(A**

**∩**

**B**

**∩**

**C**

**)**

**+ n(**

**B**

**∩**

**C**

**∩**

**D)**

**+ n(A**

**∩**

**C**

**∩**

**D)**

**- n(A**

**∩**

**B**

**∩**

**C**

**∩**

**D)**

Number of integers divisible by 2 or 3 or 5 or
7

= n(2 U 3 U 5 U 7)

= n(2) + n(3) + n(5) + n(7) – n(2∩3) – n(2∩5) – n(2∩7) – n(3∩5) – n(3∩7) – n(5∩7) + n(2∩3∩5∩7)

= 149+100+59+43-50-29-21-20-15-9+10+7+4+3-1

=

**230****Answer is 230**

__Method-II__
This is a bit trickier method saving some
amount of time:

**Number of integers divisible by 2: n(2)**

This is calculated in the similar way as in
the above method.

n(2) =

**149**

**Number of integers divisible by 3 but not by 2:**

So all even numbers between 100 and 400 are
not considered. Odd numbers divisible by 3 are considered

105,111,117,….399

This is an A.P with first term 105, last term
399 and common difference 6.

n(3 but not 2) = (399-105)/6 + 1 = 294/6 + 1
=

**50**

**Number of integers divisible by 5 but not by 2 or 3:**

The numbers divisible by 5 end with 0 or 5.
So all numbers ending with zero are not considered as they are even and
divisible by 2. And numbers ending with 5 and not divisible by 3 shall be
considered here.

Note: To eliminate the numbers divisible by
3, use 3-divisibility condition: If sum of all digits is divisible by 3, then
that number is divisible by 3

115,125,145,155,175,….385,395

Observe that the difference between two
successive numbers alters between 10 and 20 alternatively.

We can make out two sequences here:

Sequence-1:

115,145,175,….355,385

Number of terms = (385-115)/30 +1 = 10

Sequence-2:

125, 155,185,….365,395

Number of terms = (395-125)/30 +1 = 10

Total number of terms = n(5 but not 2 or 3) =
10+10 =

**20**

**Number of integers divisible by 7 but not by 2 or 3 or 5:**

This is a smaller set which can be made out
case by case:

To make it follow the rule that odd multiples
of 7 between 100 and 400 are to be considered and these numbers should not be divisible
by 2 or 3 or 5.

7*17, 7*19, 7*23, 7*29, 7*31, 7*37, 7*41, 7*43,
7*47, 7*49, 7*53

n(7 but not 2 or 3 or 5) = Total

**11**numbers.
Finally, the total number of integers =
149+50+20+11 =

**230**
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