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Senior | Next Rank: 100 Posts
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by manasi_sh » Fri Oct 05, 2007 1:12 am
How many 3 digit positive integers exist that when divided by 7 leave a remainder of 5?
A. 128
B. 142
C. 143
D. 141
E. 129

Help asap..
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by kajcha » Fri Oct 05, 2007 3:56 am
First 3 digit no that leaves remainder of 5 when divided by 7 is 103 and the last one is 999

so, total no of such numbers is = (999-103)/7 = 128

IMO ANS is A
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by manasi_sh » Fri Oct 05, 2007 4:06 am
hey thanks a ton..i'l chech d answer n let u know..can u tel me sumthg..from where du u prac math cos i m very weak at it..
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by kajcha » Fri Oct 05, 2007 4:13 am
If you think that you need to practice Maths, I would suggest you should do OG first. OG's questions are not very tough but they will give you good practice.
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by manasi_sh » Sat Oct 06, 2007 6:48 am
wel i solved OG 11 and OG 10 as well but still i m not confident at al..rather i'll be solvin KAPLAN 800 now..
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by gmatguy16 » Sun Oct 07, 2007 9:39 am
why is the answer not 129
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by jay2007 » Mon Oct 08, 2007 12:49 pm
103, 110, .... 999

Number of 3 digit numbers = ((999-103)/7)+1 = 129.

It should be 129.
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by Gunjan99 » Thu Oct 25, 2007 2:47 am
I agree with Jay2007
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