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by sparkle6 » Sun Sep 25, 2011 6:38 am
How many 3 digit positive integers exist that when divided by 7 leave a remainder of 5?

A) 128
B) 129
C) 141
D) 142
E) 143
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by knight247 » Sun Sep 25, 2011 7:02 am
The number closest to 100 that is divisible by 7 is 98. So if 5 is added to 98 and the resultant number is divided by 7 the remainder would obviously be 5. So 98+5=103 is the first 3 digit number which when divided by 7 leaves a remainder of 5. Not the greatest 3 digit number divisible by 7 is 999

Now, all the numbers which when divided by 7 leave a remainder of 5 are an evenly spaced set with an increment of 7 i.e. the difference between any two consecutive numbers in this set is 7.

For and evenly spaced set,
The number of elements=[(Biggest-Smallest)/Increment]+1=(999-103)/7+1=(896/7)+1=128+1=129
Hence B
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by amit2k9 » Sun Sep 25, 2011 7:15 am
15*7 = 105
thus 105-2 = 103 will give remainder 5.

994 is the max 3 digit number divisible by 7.

thus max number = 994+5= 999

hence
999= 103 + (n-1)*7

giving n = 129
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