What is the total number of integers between 100 and 200 that are divisible by 3?
(A) 33
(B) 32
(C) 31
(D) 30
(E) 29
PROBLEM8
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The first and last integer between 100 and 200 which are divisible by 3 are 102 and 198.Captchar wrote:What is the total number of integers between 100 and 200 that are divisible by 3?
Hence, number of integers between 100 and 200 that are divisible by 3 = number of integers between 102 and 198 (both inclusive) that are divisible by 3 = (198 - 102)/3 + 1 = 96/3 + 1 = 32 + 1 = 33
The correct answer is A.
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Lets say the question is:- What is the total number of integers between 100 and 200 that are divisible by either 3 or 4?
Then you have to take into account 3 things:-
1) Digits divisible by 3
2) Digits divisible by 4
3) Digits divisible by 12, as these have been counted twice in calculations above.
So Answer becomes (1) + (2) - (3)
i.e. (1) becomes (198 - 102)/3 + 1 = 33
(2) becomes (200 - 100)/4 + 1 = 26
(3) becomes (192 - 108)/12 + 1 = 8
Answer is 33 + 26 -8 = 51.
Hope this makes it clear!!!
Then you have to take into account 3 things:-
1) Digits divisible by 3
2) Digits divisible by 4
3) Digits divisible by 12, as these have been counted twice in calculations above.
So Answer becomes (1) + (2) - (3)
i.e. (1) becomes (198 - 102)/3 + 1 = 33
(2) becomes (200 - 100)/4 + 1 = 26
(3) becomes (192 - 108)/12 + 1 = 8
Answer is 33 + 26 -8 = 51.
Hope this makes it clear!!!
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And Lets say if the question is:- What is the total number of integers between 100 and 200 that are divisible by both 3 and 4?
So we have to look for digits divisible by 12(LCM of 3 and 4) only.
This equals (192 - 108)/12 + 1 = 8.
Hope this helps!!!
So we have to look for digits divisible by 12(LCM of 3 and 4) only.
This equals (192 - 108)/12 + 1 = 8.
Hope this helps!!!
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We know that the multiples of 3 are 102, 105, 108, . . . 195, 198, and we want to know how many numbers there are in this sequence.Captchar wrote:What is the total number of integers between 100 and 200 that are divisible by 3?
(A) 33
(B) 32
(C) 31
(D) 30
(E) 29
Some people will use the formula for arithmetic progressions here, but I think that we can solve this kind of question just as fast (and without needing to memorize formulas) by looking for a pattern.
Notice that:
102 = (3)(34)
105 = (3)(35)
108 = (3)(36)
111 = (3)(37)
.
.
.
195 = (3)(65)
198 = (3)(66)
So, the question becomes "How many integers are between 34 and 66 inclusive?"
66 - 34 + 1 = 33, so there are 33 integers are between 34 and 66 inclusive, which means there must be 33 integers between 100 and 200 that are divisible by 3.
Answer = A
Cheers,
Brent
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You can also solve this problem using the floor function, [200/3] - [100/3] = 66-33 = 33 (where [] denotes the floor function)Captchar wrote:What is the total number of integers between 100 and 200 that are divisible by 3?
(A) 33
(B) 32
(C) 31
(D) 30
(E) 29
The biggest advantage here is that you need not bother about the boundary values or need not apply the inclusion exclusion principle.
PS:The floor function or the greatest integer function gives the largest integer less than or equal to the number. e.g. [11.9]=11, [-2.4]=-3, [3]=3
So essentially you can find the number of multiples of any number less than a particular number.
e.g. number of multiples of 7 upto 200 = [200/7] = 28
Let's solve another problem applying this function. Let's say we need to find how many numbers upto 400 are divisible by 3 and 5.
[400/3] + [400/5] - [400/15] = 133+80-26=187
Note that we found all the multiples of 3, all the multiples of 5 and substracted the overlapping multiples which are both the multiples of 3 and 5.
Another application of this function is to calculate the highest power of a prime number p in n!
[n/p]+[n/p^2]+[n/p^3]+[n/p^4]+... where [] denotes the floor function.
Let's aplly this to a problem.
Find the number of positive divisors of 15!.
The prime factors in 15! are 2,3,5,7,11, and 13.
Powers of 2 in 15! = [15/2]+[15/2^2]+[15/2^3]=7+3+1=11
Powers of 3 in 15! = [15/3]+[15/3^2]=5+1=6
Powers of 5 in 15! = [15/5]=3
Powers of 7 in 15! = [15/7]=2
Powers of 11 in 15! = [15/11]=1
Powers of 13 in 15! = [15/13]=1
Therefore 15!=(2^11)*(3^6)*(5^3)*(7^2)*11*13
Number of positive divisors = 12*7*4*3*2*2= 4032