## Sunday, 18 December 2011

### 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.
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 = tn = 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(AB) – n(AC) – n(AD) – n(BC) – n(BD) – n(CD) + n(ABC+ n(BCD) + n(ACD) - n(ABCD)
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(23) – n(25) – n(27) – n(35) – n(37) – n(57) + n(2357)
= 149+100+59+43-50-29-21-20-15-9+10+7+4+3-1
= 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