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Difficult Math Question #51 - Arithmetic

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Difficult Math Question #51 - Arithmetic

by 800guy » Fri Nov 10, 2006 5:08 pm
How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3?

A.14
B.15
C.16
D.17
E.18
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by rajs.kumar » Sat Nov 11, 2006 3:53 am
Answer is 17.

# terms in the arithmetic progression from 1 to 49 with a difference 3 (inclusive)

49 - 1/3 + 1 = 17
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OA

by 800guy » Mon Nov 13, 2006 4:58 pm
OA:

if we arrange this in AP, we get
4+7+10+.......+49

so 4+(n-1)3=49: n=16
C is my pick
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Re: OA

by rajs.kumar » Tue Nov 14, 2006 4:51 am
800guy wrote:OA:

if we arrange this in AP, we get
4+7+10+.......+49

so 4+(n-1)3=49: n=16
C is my pick
It is division and not modulo so I was wrong in including 1. With that correction I will get 16 as well.

Thanks 800
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by Umar82 » Fri May 01, 2009 12:37 pm
Guys,

The answer is 17. Remember that 1 divided by 3 also leaves a remainder of 1 . . . so we have

1,4,7,10,13,16,19,22,25,28,31,34,37,40,43,46,49
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by abhinav85 » Fri May 01, 2009 12:43 pm
Please correct me if i am wrong?

How u can get remainder 1 when 1 is divided by 3??

Can u explain it? may be i am missing something?
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by awesomeusername » Fri May 01, 2009 1:12 pm
1/3 = 0R1

1 goes into 3, 0 times
1 minus 0 = 1.

I thought it was 17 too.
Constant dripping hollows out a stone.
-Lucretius
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by maihuna » Fri May 01, 2009 1:25 pm
3x+1 from 0 to 50 like 1, 4, ..., 49
a total of 17 for x=0 to 16
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