How many positive integers between 200 and 300 (both inclusive) are not divisible by 2, 3 or 5?
A. 3
B. 16
C. 75
D. 24
E. 25
Is there any quicker way to find out except writing down the no.s?
source 4Gmat.com and OA is 25...
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- kinjal.nandy
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- JonathanSchneider
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Re: Divisibilty prob
Well, I've solved this two ways and both times shown the answer to be 26. Maybe I'm missing something?
My first thought:
There are 101 numbers in this range. Only 50 of these numbers, however, are odd. 51 are even. The evens are out, as they are divisible by 2.
Of the remaining 50, 1/3 of these will be divisible by 3. 50/3 = 16 + 2/3. So, we must decide, does this mean that 16 of the remaining numbers are divisible by 3, or is it 17? Well, the 2/3 really means that we have a remainder of 2, or two extra numbers that are not in a perfect set of three. As the first remaining number, 201, is divisible by 3, we know that the extra two numbers includes a number that is divisible by 3. Thus, 17 of our remaining 50 are divisible by 3; these are now out. 33 remain.
Of the remaining 33, one fifth will be divisible by 5. 33/5 = 6 + 3/5. Again, we must decide, do we have 6 multiples of five, or 7? As the first multiple of 5 remaining, 205, appears within the first three numbers, we have 7 in all. Thus, 7 numbers are now ruled out. This leaves us with 26.
My second approach was to create a chart. Here I lined up 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 along the top, to signify the units digit. Then I lined up 200, 210, 220, 230, 240, 250, 260, 270, 280, and 290 along the side, to signify the tens digit. (We do not need to consider 300, as it is clearly divisible by 2, 3, and 5.) From here, I scratched out the entire columns labeled 0, 2, 4, 5, 6, and 8. The evens are out, and so are those that end in 5. This leaves us with only those columns that start with 1, 3, 7, and 9. Now, we must determine how many numbers will be ruled out from each column, but determining how many multiples of three lie in each. 201 is a multiple of 3 (its digits add to 3), so that is out. The next number in that column to rule out is simply 30 more, or 231. 261 and 291 are also out. We have 6 that remain. From the "3" column, we have 7 that remain (this can be seen more quickly because there are only three multiples of three here). Ditto for the "7" column. The "9" column has six that remain. In all, we have 6+7+7+6 = 26 that remain.
I'm not sure either is a realistic two-minute method, though the second one was probably pretty close.
My first thought:
There are 101 numbers in this range. Only 50 of these numbers, however, are odd. 51 are even. The evens are out, as they are divisible by 2.
Of the remaining 50, 1/3 of these will be divisible by 3. 50/3 = 16 + 2/3. So, we must decide, does this mean that 16 of the remaining numbers are divisible by 3, or is it 17? Well, the 2/3 really means that we have a remainder of 2, or two extra numbers that are not in a perfect set of three. As the first remaining number, 201, is divisible by 3, we know that the extra two numbers includes a number that is divisible by 3. Thus, 17 of our remaining 50 are divisible by 3; these are now out. 33 remain.
Of the remaining 33, one fifth will be divisible by 5. 33/5 = 6 + 3/5. Again, we must decide, do we have 6 multiples of five, or 7? As the first multiple of 5 remaining, 205, appears within the first three numbers, we have 7 in all. Thus, 7 numbers are now ruled out. This leaves us with 26.
My second approach was to create a chart. Here I lined up 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 along the top, to signify the units digit. Then I lined up 200, 210, 220, 230, 240, 250, 260, 270, 280, and 290 along the side, to signify the tens digit. (We do not need to consider 300, as it is clearly divisible by 2, 3, and 5.) From here, I scratched out the entire columns labeled 0, 2, 4, 5, 6, and 8. The evens are out, and so are those that end in 5. This leaves us with only those columns that start with 1, 3, 7, and 9. Now, we must determine how many numbers will be ruled out from each column, but determining how many multiples of three lie in each. 201 is a multiple of 3 (its digits add to 3), so that is out. The next number in that column to rule out is simply 30 more, or 231. 261 and 291 are also out. We have 6 that remain. From the "3" column, we have 7 that remain (this can be seen more quickly because there are only three multiples of three here). Ditto for the "7" column. The "9" column has six that remain. In all, we have 6+7+7+6 = 26 that remain.
I'm not sure either is a realistic two-minute method, though the second one was probably pretty close.
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