Series manipulation

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Series manipulation

by rnaah » Fri Sep 30, 2011 2:45 am
In the series a1+a2+a3+....+an each term is either 7 or 77 and the sum is 350. which of the following could be the value of n? 38, 39, 40, 41, and 42.
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by Brent@GMATPrepNow » Fri Sep 30, 2011 4:06 am
rnaah wrote:In the series a1+a2+a3+....+an each term is either 7 or 77 and the sum is 350. which of the following could be the value of n? 38, 39, 40, 41, and 42.
Looking for a pattern is one possible route:

Since both 7 and 77 have 7 as their units digit, we know that if we take any two terms, their sum will have a units digit of 4 (e.g., 7 + 7 = 14, 7 + 77 = 84, 77 + 77 = 154)

Similarly, if we take any three terms, their sum will have a units digit of 1. (e.g., 7 + 7 + 7 = 21, 7 + 7 + 77 = 91, etc.)

Now let's look for a pattern.

The sum of any 1 term will have units digit 7
The sum of any 2 terms will have units digit 4
The sum of any 3 terms will have units digit 1
The sum of any 4 terms will have units digit 8
The sum of any 5 terms will have units digit 5
The sum of any 6 terms will have units digit 2
The sum of any 7 terms will have units digit 9
The sum of any 8 terms will have units digit 6
The sum of any 9 terms will have units digit 3
The sum of any 10 terms will have units digit 0
The sum of any 11 terms will have units digit 7 (at this point, the pattern repeats)

From this, we can conclude that the sum of any 20 terms will have units digit 0
And the sum of any 30 terms will have units digit 0, and so on.

We are told the sum of the terms is 350 (units digit 0), so the number of terms must be 10 or 20 or 30 or . . .

Since C is a multiple of 10, this must be the correct answer.


Another option is to try some different configurations. If we do, we see that adding 39 7's and 1 77 gives us a total of 350. (39 + 1 = 40)

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by GmatMathPro » Fri Sep 30, 2011 8:42 am
This is similar to Brent's second solution, but you could also reason that 50 7's would give you 350. For each 77 that you add in, you have to remove 11 7's to keep the sum at 350. If we're looking for something in the range of 38-42 total terms, we can only afford to add in 77 once. This would decrease your total number of terms by 10 (plus one term for the 77 and minus eleven terms for the removed 7's. So you have [spoiler] 50-10=40[/spoiler] terms.
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by phoenix111 » Fri Sep 30, 2011 10:40 pm
rnaah wrote:In the series a1+a2+a3+....+an each term is either 7 or 77 and the sum is 350. which of the following could be the value of n? 38, 39, 40, 41, and 42.
Let number of 7's be X and number of 77's be Y:

n = X + Y

7X + 77Y = 350
or
X + 11y = 50
Possible values of X,Y : (39,1)(28,2)(17,3)(6,4)
First one gives X + Y = 40

So answer: 40