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Sequence Problem

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by oex85 » Tue Aug 09, 2011 6:43 pm
I'll try for practice - hope you don't mind (and that I'm correct!).

Lets start with all 7's: 350 (7x5x5x2) = 50x7. Since 50 is not an alternative, lets look at reducing the number of 7's and including 77's.

77 = 7x11. This gives us a hint that we should only have ONE 77 in the sequence to end up among the answer choices.

50-11=39. But don't forget to add the 77: 39+1=40.

Would be interested in seeing other approaches (assuming my answer is correct)!
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by ashutoshkumar7 » Tue Aug 09, 2011 8:25 pm
Since the sum's last digit is zero so it means that the n is a multiple of 10.
if n had been 2 then last digit would have been 4, if 3 then 1'and so on
So the answer is 40 by looking at the choices.
Thanks,
Ashutosh
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by Anurag@Gurome » Tue Aug 09, 2011 8:27 pm
Spartacus2000 wrote:If each term in the sum A1 + A2 + ..... + An is either 7 or 77 and the sum equals 350, which of the following could be equal to n?
Since the units digit of the sum is zero, n must be such that adding 7 n times will result a units digit of 0.

Only for n = 40, the units digit will be zero.

The correct answer is C.
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by krishna8143 » Tue Aug 09, 2011 9:22 pm
ans is 40

350 can be written as 7+7+....... 50 times

if there is only 7 and 77 are the numbers then by looking at the options we can tell there is only one 77 exists

so remaining 39 numbers are 7s

so the total numbers are 40
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