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21 Oct 2007, 20:50
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How many 3 digits numbers are there so that each is evenly divisible by 7?
105
111
127
128
142
Do we use AP in this?
105
111
127
128
142
Do we use AP in this?
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21 Oct 2007, 21:40
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bmwhype2
D.
Last term = a + (n-1)d where a is the 1st term and d is the common difference.
1st 3 digit integer divisible by 7 = 105
last 3 digit integer divisible by 7 = 998
998 = 7 + (n-1)7
Solve for n = 128.
