If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum

7x + 77(n-x) = 350 --> where x is the number of times 7 repeats
7x + 7(11)(n-x) = 350
dividing both sides by 7
x + 11 (n-x) = 50

trying different options

now (n-x) has to be 1 because if its more than 1 (ex. 2) then 11(n-x) = 22 and x will be 37 or more which takes the total beyond 50.

Therefore now trying options we get

38 --> 37 + 11(1) = 48
39 --> 38 + 11(1) = 49
40 --> 39 + 11(1) = 50 ... this is the right answer

This is as good as saying all the numbers are either 1 or 11 and the sum equals 50
Let 1s be x and 11s be y, thus making n = x+y
x + 11y = 50
(x+y)+10y = 50
x+y = 10(5-y)
Hence x+y must be a multiple of 10 i.e. n must be multiple of 10. Only choice C

In case if the question asks the possible values of n, we can conclude that n could be 10,20,30,40,50

\(\frac{350}{7} = 50\);

As all the options are below 50, there should be atleast one 77 present

To add one 77, we require to remove eleven 7's (7 x 11)

50 - 11 + 1 = 40 = Answer = C

If 40 wasn't there, then again same method:

40 - 11 + 1 = 30

Hi,
One other approach could be this.
350/7 =50.
Implies, it takes 50 7s or 0 77s to get a 350.
Considering 77 is 11 7s put together, try 1 77 and 39 7s (50-11 7s).
There u go. 1+39 = 40 !

Same concept alternate solution

Let the number of 77's be X, we know \(X>=0\)

\(7*(n-X) + 77*X = 350\)
\(7*(n-X) + 7*X + 70*X = 350\)
\(7*n + 70*X = 350\)

Since \(X>=0\), \(X\) can be \(0,1,2,3,4,5\) .... Correspondingly the value of n would be \(50,40,30,20,10,0\) (always a multiple of 10)

So the answer in this case is 40.

Consider this 7+7+7+7+7+7+7+7+7+7=10*7=70 (the sum of ten 7's equals to 10 times that term).

Hope it helps.

350= 7a + 77b (a= no of 7s & b = no of 77s)

If b=0 then a=50
If a=0 then b=4 (+some remainder)

both these answers are not in the options.And looking at the options and above range we can conclude that the sum contains atleast 1 term as 77

put b=1,

350= 7a + 77(1)
350 - 77 = 7a

50-11 = a

a=39

a+b = 39+1 = 40 which is in the options

hence C

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 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 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 = 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

Check the solution in attachment

Given \(A1 + A2 + ... An = 350\)
Let there be \(x\) 7 and \((n-x)\) 77 in the set.

so \(x * 7 + (n-x) * 77 = 350\)
= \(7 * (x + (n-x)*11) = 7* 50\)
= \((x + 11n - 11x) = 50\)
= \(11n - 10x = 50\)
Since right hand side is a multiple of 10,then n has to be a multiple of 10 for integer \(x\) value and only C)40 . \(( x = (11 n - 50)/10)\)


+1 for kudos

Solution:

Let there be a number of 7s and b number of 77s

=> Total number of terms = a + b

=>7a + 77b = 350

=> a + 11b = 50

Its a Diophantine equation and we can solve it by checking for values of a and b

If b=1 then a should be 39 and total terms = a +b =1+39 =40

=> Total terms is present among the options

Hence 39 (option c)

Devmitra Sen
GMAT SME

Thank you, Bunuel. Your first solution is a good one (the second one is rather non-deterministic for me).

Thanks Bunuel for both the solutions. Very much appreciated.

Answer - C)

Let the no. of 7s be a and the no. of 77s be b

7a + 77b = 350

7(a+11b) =350

a+b+10b = 50

As a+b = n, n = 50-10b (50- a multiple of 10)

The only ans choice which is a multiple of 10 is C)40

Hi Bunuel,

Thanks for all your help, as usual. I answered the question using the second approach but would love to understand the first approach better. I must be missing something simple, but could you further explain why the number off terms must be a multiple of 10 to get a "0" in the units digit? That statement is tripping me up.

We will start with the bigger numbers. There can be maximum of 50 Sevens and one 77 will replace 11 Sevens.

For there to be 50 terms: there will be One 77 (11 sevens) and 39 sevens. Hence 39 + 11 = 50 sevens.

Hence the answer is C.

Hi Bunuel,

I noticed that you tried to clarify method 1 below but still not connecting.

How are you drawing the conclusion that it has to be a multiple of 10? Why can't it be 2*175 which is NOT a multiple of 10 or 1*350 etc?

Thanks!