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Number of terms in a sum

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by amirhakimi » Wed Nov 06, 2013 9:07 am
If each term in sum a1+a2+ ... +an is either 7 or 77 and the sum equals 350, which of the following could be equal to n?

A)37
B)39
C)40
D)41
E)42

Answer is C
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by Brent@GMATPrepNow » Wed Nov 06, 2013 9:09 am
amirhakimi wrote:If each term in sum a1+a2+ ... +an is either 7 or 77 and the sum equals 350, which of the following could be equal to n?

A)37
B)39
C)40
D)41
E)42
Notice that 77 does not divide into 350 many times.
In fact, there can be, at most, four 77's in the sum
So, there are only 5 cases to consider (zero 77's, one 77, two 77's, three 77's and four 77's)
It shouldn't take long to check the cases.

case 1: zero 77's in the sum
If every term is 7, the total number of terms is 50.
50 is not one of the answer choices, so move on.

case 2: one 77 in the sum
350 - 77 = 273
273/7 = 39
So, there could be thirty-nine 7's and one 77 in the sum, for a total of 40 terms.

This matches one of the answer choices, so the correct answer is C

Cheers,
Brent
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by Brent@GMATPrepNow » Wed Nov 06, 2013 9:11 am
amirhakimi wrote:If each term in sum a1+a2+ ... +an is either 7 or 77 and the sum equals 350, which of the following could be equal to n?

A)37
B)39
C)40
D)41
E)42
Another possible approach it to look for a pattern:

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.

Cheers,
Brent
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by gmatclubmember » Wed Nov 06, 2013 10:30 am
Let there be 'a' 7s and 'b' 77s.
7a+77b=350 => a+11b=50.
Case1: when b=1, then a=39, then n = 40
Case2: when b=2, then a=28, then n=30
Case3: when b=3, then a=17, then n=20
Case4: when b=4, then a=6 ,then n=10

Possible solution is n=40.
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by gmatclubmember » Wed Nov 06, 2013 10:43 am
If you really want to solve the above mentioned equation (in my post above) there is a solution as mentioned below:
We know that a and b can be integer (cant be fractions)
So, a+11b=50 => 11b=50-a => b=4+(6-a)/11
LHS and RHS should both be integers (since LHS is an integer).
So it means that (6-a)/11 = p (lets say p is an integer)
which gives us a=6-11p ----------(i)
Substitute this in a+11b=50 => b=4+p ---------(ii)
since p is an integer it can take -1, -2, -3, and 0 values only. Any other value of p will give a or b negative which is not possible. So the no. of terms in each of the valid cases would correspond to 20, 30, 40 and 10.

This way of solving one equation in 2 variables will work only if the variables are integer.
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