There are n terms, and each term is either 7 or 77. This means if we factor out 7 from the expression, we have either 1 or 11 in parenthesis.

We can therefore reduce the problem to:

Each term in the sum b1 + b2 + b3+...+bn is either 1 or 11, and the sum equals 50. What could n be?

Let's start by assuming there are no 11's. This means there must be 50 1's => n= 51 (Not in the choice)
Continuing on this trend:

1 eleven and 39 1's n = 40 (In our choice! Pick it )

Pick C

7a+77b=350

a+11b=50

Plug in the options in the above equation,we will get a =39 and b=1

n=40

If each term in the sum a1 + a2 + a3...+ an is either 7 or 77 and the sum equals 350, which of the following could be n?

A) 38
B) 39
C) 40
D) 41
E) 42
350 if you divide by 7 you will get 50 so you can havefifties 7 but we need to include 77 also
now let us include one 77
so we can have 350-77=273/7= 39 7's
so n could be 40

now let us take 2 77's now number of 7 will be (350-(77*2))/7=28 7's so n will be 30 in that case
so from given option we can conclude answer will be 40
cheers

san2009 wrote:Please provide a conceptual/methodological approach only

If each term in the sum a1 + a2 + a3...+ an is either 7 or 77 and the sum equals 350, which of the following could be n?

7+77+77+...
Let number of 7's be x and 77's be y

7x+77y = 350
x+11y= 50, both x and y are positive integers.
y = (50-x)/11 should be an integer. So 50-x should be multiple of 11.

50-x = 11 -> x =39, y = 1
50-x =22 -> x=28, y=2
50-x = 33 -> x = 17, y=3,
50-x = 44 -> x = 6, y=4

we need x+y so answers are, 40,30,20,10