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sum a1+a2+a3...+an

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sum a1+a2+a3...+an

by spider » Tue Apr 10, 2012 12:26 am
XCan anyone help me with this sum.

If each term in the sum a1+a2+a3...+an is either 7 or 77 and the sum is 350, which of the following could be n?
a)38
b)39
c)40
d)41
e)42
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by Anurag@Gurome » Tue Apr 10, 2012 12:40 am
spider wrote:XCan anyone help me with this sum.

If each term in the sum a1+a2+a3...+an is either 7 or 77 and the sum is 350, which of the following could be n?
a)38
b)39
c)40
d)41
e)42
Let x and y be the number of 7's and the number of 77's respectively.
Then 7x + 77y = 350 implies x + 11y = 50
x + y = 50 - 10y
So, x + y = n = 50 - 10y implies n must have the unit digit of 0 because 50 - 10y must be a number which has units digit as 0. C is the only option in which the units digit is 0.

The correct answer is C.
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by Shalabh's Quants » Tue Apr 10, 2012 1:27 am
spider wrote:XCan anyone help me with this sum.

If each term in the sum a1+a2+a3...+an is either 7 or 77 and the sum is 350, which of the following could be n?
a)38
b)39
c)40
d)41
e)42

We know that unit digit in each term is 7. We also know that the sum's unit digit is 0. Adding few 7s and getting 0 as unit digit is possible if the number of terms is a multiple of 10.

Only in option (c) 40 is a multiple of 10.
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by GMATGuruNY » Tue Apr 10, 2012 3:12 am
spider wrote:XCan anyone help me with this sum.

If each term in the sum a1+a2+a3...+an is either 7 or 77 and the sum is 350, which of the following could be n?
a)38
b)39
c)40
d)41
e)42
ALWAYS LOOK AT THE ANSWERS.

50*7 = 350, implying that 50 terms of 7 will yield a sum of 350.
Since all of the answer choices are less than 50, the sequence must include AT LEAST one 77.
If there is one 77:
350-77 = 273.
273/7 = 39.
This works: 1(77) + 39(7) = 350.
Thus, the total number of terms = 1+39 = 40.

The correct answer is C.
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