Fab_Vr6 wrote:If each term in the sum a1+a2+.....an (n-term) is either 7 or 77, and the sum is equal to 350, which of the following could equal to n?

1) 38
2)39
3)40
4)41
5)42

You could work from answer choices. You will find that 40 would be the answer. The sequence would be that it has 39 -7s and 1 -77, such that the sum=(39*7)+77=273+77=350.

but how would you know when using 40, to start with 39 7's and 1 77? from where should you start looking?

When I looked at this problem I didn't look at the answers very closely(big mistake,...I know). I divided 77 into 350 and saw that it went in 4 times with a remainder of 42, which is 6 sevens.

That only gave me n=10, which is the smallest n could be. Then I looked at the answers and realized that in order to be up around 40, there would have to be alot more sevens than seventy-sevens.

So I subtracted seventy seven from 350 leaving 273, and tried dividing that by 7. It went 39 times, leaving me with n=40 which was an available answer.

You can quickly see the other anwers aren't viable by realizing if you have two 77s, n is no where near high enough, and all 7s would be n=50.

I'm sure there is a more "scientific" way to attack it, but that was my approach. Hope it helps.

Assume that the number of 7s and 77s in the series is x and y respectively.

Then we could form the equation:

7x + 77y = 350
=> 7 (x + 11y) = 350
=> x + 11y = 50

Possible values of (x,y) : (39,1) , (28,2) , (17,3) .. and so on (you just need to calculate with y = 1)

x + y is the total number of terms in the series i.e n = x + y
n can take values : (40, 30, 20) .. and so on

so 40 is the answer ...

thanks guys!!!