How many #s between 18 and 102, inclusive, are multiples of

Question

How many #s between 18 and 102, inclusive, are multiples of both 2 and 3? a) 18 b) 17 c) 16 d) 15 e) 14 OA to follow.

:

sumande · Answer

I think it should be D.

If the number is divisible by both 2 and 3, it must be divisible by 6.
Therefore, number of such numbers (102-18)/6 + 1 = 15

FN · Answer

its D and here is why


102-18=84 so the LCM of 2 and 3 is 6

84/6 = 14

ok but we got to check for 36 also, 84/36= 1 whole number + a frac

totall number of multiples of 6 are 14+1=15

zakk · Answer

c. 16

*edit* I was wrong, don't want to mess anyone up.

ywilfred · Answer

Number must be a multiple of both 2 and 3, so the number must be a multiple of 6.

18 is the smallest multiple of 6
102 is the largest multiple of 6

# of multiples of 6 = (102-18)/6 + 1 = 15

CaspAreaGuy · Answer

D. 
2*3=6
102/6 = 17 numbers
substract 2 multiples of 6 that are not in the range to get 17-2=15

Dellin · Answer

15 is right 
divisible by both 2 and 3, the should be divisible by 6
18= 3x6 and 102 = 17x6 since both nos are inclusive
17-3+1 = 15

buffdaddy · Answer

D

here's how i got there

Find prime factors of both 18 and 102

i.e 18 --> 2*3 * 3
 102 --> 2*3 * 17

so you can see 2*3 = 6. so the question is actually asking howmany integers between 3 and 17 inclusive.

which is 15

How many #s between 18 and 102, inclusive, are multiples of

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