How many positive integers less than 100 are neither multiples of 2 or

Hi,

To answer this Q we require to know

1) multiples of 2 till 100 \(= \frac{100}{2} = 50\)
2) Multiples of 3 till 100 = \(\frac{100}{3} = 33.33= 33\)
add the two \(50+33=83\) ; subtract common terms that are multiple of both 2 and 3..

LCM of 2 and 3 = 6
Multiples of 6 till 100 = \(\frac{100}{6} = 16.66 = 16\)
so total multiples of 2 and 3 = 83-16 = 67

ans = \(100-67 = 33\)

D
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Edited Solution:

The question states positive integers less than 100. i.e. 1, 2, 3, .... 99

Number of multiples of 2 less than 100 = 49 (2*49 = 98)
Number of multiples of 3 less than 100 = 33 (3*33 = 99)

Some of the integers that are divisible by both 2 and 3 are double counted.
LCM(2, 3) = 6
Number of multiples of 6 less than 100 = 16 (6*16 = 96)

Number of positive integers that are not divisible by 2 or 3 = 99 - (49 + 33 - 16) = 100 - 66 = 33

Answer: D

So how I approached that question with sequences technique.

1) Total numbers in the set: \(\frac{(99-1)}{1}\) + 1=99

2) Find # of multiples of 2: \(\frac{(98-2)}{2}\) +1=49

3) Find # of multiples of 3: \(\frac{(99-3)}{3}\) +1=33

4) (There is an intersection-overlap between the two above) so find # of multiples of 6: \(\frac{(96-6)}{6}\) +1=16

5) All the rest: 99 - [(49+33)-16] = 33

To find the respective multiples I narrow the range accordingly by shifting the upper and lower limits to enclose the exact relevant multiples. Say 98 is the largest mulptiple of 2 in the given set from 1 to 99. Though neither 99 nor 98 is a multiple of 6 so I shifted down do 96 at the upper limit and raised the lower limit to the first available multiple of 6 that is 6 itself.

Hi Vyshak,
Since 100 is being negated as div by 2, it did not make any difference if we take it or not..
But if we are not taking it, as correctly observed by you, DO not take in total--


Here 100 should be 99, and answer will be 99-66 = 33
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Multiples 2 and 3 is ( 50 + 33 ) = 83

Multiple of 6 = 16

While counting the multiples of 2 and 3 we have already included the multiple of 6 ( Which itself is a multiple of 2 & 3 ) so we need to subtract it from the multiples of 2 & 3 to reach the correct answer.

So, the total multiple of 2 and 3 is 67 ( 83 - 16 )

Hence answer will be (D) 67
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Multiples of 2 = (98 - 0)/2 + 1 = 50
Multiples of 3 = (99 - 0)/3 + 1 = 34
Multiples of 6 = (96 - 0)/6 + 1 = 17

Num of multiples = 50 + 34 - 17 = 67

Numbers that are not multiples of 2 and 3 = 100 - 67 = 33. D


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hi although you reach the right answer your calculation is not accurate and may play bad with you on other questions - that is you will get a wrong answer. The set is exclusive of 0 as the stem clearly limits the numbers to positive ones only. Look at my solution it is slick neat and beautiful lol =). To find the respective multiples I narrow the range accordingly by shifting the upper and lower limits to enclose the exact relevant multiples. Say 100 is a mulptiple of 2 that is why I left it as it is. Though 100 is not the multiple of 6 so I shifted down do 96 at the upper limit and raised the lower limit to the first available multiple of 6 that is 6 itself.

Hello,

My method is not wrong, but I did make a mistake of including 0 as the lower limit as I misread the question. But looking at the math again, the answer should be 34. As the question stem clearly states 'less than 100' and does not imply that 100 is inclusive.

yeah man u are right i also made the same mistake! i erroneously included 100 while i should not have. i ve edited my solution. see now

Correcting my solution:

Multiples of 2 = (98 - 2)/2 + 1 = 49
Multiples of 3 = (99 - 3)/3 + 1 = 33
Multiples of 6 = (96 - 6)/6 + 1 = 16

Numbers between [1-99] = 99

Num of multiples = 49 + 33 - 16 = 66

Numbers that are not multiples of 2 and 3 = 99 - 66 = 33. D


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Numbers less than 100 that are neither multiples of 2 or 3.

Multiples of 2 = 2, 4, ... 98 - 49 numbers
Multiples of 3 = 3, 6, 9, ... 99 = 33
Multiples of 6 - 6, 12, ... 96 - 16numbers

Hence total numbers that are either multiples of 2 or 3 = 49 + 33 - 16 = 66
Hence numbers that are not multiples of 2 or 3 = 99 - 66 = 33

Correct Option: D

Two ways this question can be approached =>
multiples of 2=>50
Multiples of 3=> 33
Multiples of 6=> 16(BOTH)
either 2 and 3 multiples => 50+33-16=> 77
Neither nor (2,3) => 100-77=> 33

Another approach can be this =>
Non multiples of 2=> 50
multiples of 3 out of these => 3,9,15....99=> 17
hence leftovers => 50-17=> 33


Smash that D
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Bro, although you got the answer correct , you have missed "positive integers less than 100".

So, we need to consider the positive integers from 1 to 99 only.
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99-98/2-99/3+96/6=33

Oops ..!!
I indeed missed that.
Thanks man


+1 to you
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I went about this problem by

first, calculating how many times 2 goes into 99 --> 49

second, "..." 3 goes into 99 --> 33

third, "..." 6 goes into 99 --> 16

Overall solution will be: 100-(49+33)+16 =33 --> the reason we add back in the 16 is because we have already counted multiples of 6 when we calculated 2 and 3 into 99.

D.

1)Multiples of 2: (98-2)/2+1=96/2+1=48+1=49 2)Multiples of 3: (99-3)/3+1=32+1=33 3)Multiples of 2&3: (96-6)/6+1=16 3)Positive integers less than 100 that are neither multiples of 2 or 3 is 99-49-33+16=33

Hi chetan2u,

thank you for the reply, what if we were asked to find out the number of multiples of 7 between and including 270 and 500?

Considering both 270 & 500 are inclusive,

The first multiple of 7 after 270 = 273 (took 273 because 270 itself is NOT a multiple of 7)
The last multiple of 7 before 500 =497 (took 497 because 500 itself is NOT a multiple of 7)

Problem boils down to finding the number of terms in Arithmetic progression starting 273 and ending 497.

Use tn = a + (n-1)d

tn= 497 ; a=273 ; d =7. Solve for n. n = 33.

So 33 terms.