LM wrote:
If each term in the sum \(a_1+a_2+a_3+...+a_n\) is either 7 or 77 and the sum equals 350, which of the following could be equal to n?
A. 38
B. 39
C. 40
D. 41
E. 42
APPROACH #1: 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 sumIf 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 sum350 - 77 = 273
273/7 = 39
So, there could be 39 7's and 1 77 in the sum, for a total of 40 terms
This matches one of the answer choices, so the correct answer is C
APPROACH #2: Another possible approach is 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 = 1
4, 7 + 77 = 8
4, 77 + 77 = 15
4)
Similarly, if we take any
three terms, their sum will have a units digit of 1. (e.g., 7 + 7 + 7 = 2
1, 7 + 7 + 77 = 9
1, 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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