Numbers – 22

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