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from 0 to 50, inclusive

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by sanju09 » Thu Mar 26, 2009 6:20 am
How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3 ?
A. 15
B. 16
C. 17
D. 18
E. 19
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by scoobydooby » Thu Mar 26, 2009 6:29 am
the numbers are of the form: 3k+1

1st number: 3.0+1=1
2nd number: 3.1+1=4
3rd number: 3.2+1=7
number closest to 50 divisible by 3=48, so the last number is 49 (1 more than a multiple of 3)

we have a series 1+4+7+..+49... (an AP, with a=1, d=3, Tn=49)
Tn=a+(n-1)d
49=1+(n-1)3
48/3=16=n-1
or n=17

C
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by truplayer256 » Thu Mar 26, 2009 2:08 pm
I'm going with B.
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Re: from 0 to 50, inclusive

by Stuart@KaplanGMAT » Thu Mar 26, 2009 2:48 pm
sanju09 wrote:How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3 ?
A. 15
B. 16
C. 17
D. 18
E. 19
When you divide by any number n, there are n possible remainders. For example, when we divide by 3, the 3 possible remainders are 0, 1 and 2. When we divide by 4, the 4 possible remainders are 0, 1, 2 and 3.

So, in this question there are 3 possible remainders.

From 0 to 50 inclusive there are 51 numbers. 51/3 = 17, therefore there are 17 numbers in the set that give each remainder. Finito, choose (C).

If 3 didn't divide evenly into 51, then we'd have to see if we had a "bonus" number in the set; but on this particular question, we see that 51 is evenly divisible by 3 and, if we understand the underlying concepts, get the correct answer in approximately 14 seconds!
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