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

Question

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

Bunuel · Accepted Answer

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

Number of approaches are possible.

For example, approach #1:

Since the units digit of 350 is zero then the number of terms must be multiple of 10. Only answer choice which is multiple of 10 is C (40).

To illustrate consider adding:

*7
*7
...
77
77
----
=350

So, several 7's and several 77's, note that the # of rows equals to the # of terms. Now, to get 0 for the units digit of the sum the # of rows (# of terms) must be multiple of 10. Only answer choice which is multiple of 10 is C (40).

Answer: C.

Approach #2:  
 
 7x+77y=350 , where  x  is # of 7's and  y  is # of 77's, so # of terms  n  equals to  x+y ;

7(x+11y)=350  -->  x+11y=50  --> now, if  x=39  and  y=1  then  n=x+y=40  and we have this number in answer choices.

Answer: C.

Hope it helps.

MHIKER · Answer

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

350 \ can \ be \ comprised \ of \ 7 \ but \ not \ of \ 77. \ As \ 350 \ is \ not \ a \ multiple \ of \ 77.

For example  \frac{350}{7}=50 , but  \frac{350}{77}=4.54..  is not an integer

So  50   can be an answer but it is not in the answer choices.

Thus \ we \ infer \ that \ there \ must \ be \ at \ least \ one \ 77 \ in \ 350

Let's \ Check

350 - 77 = 273

\frac{273}{7}= 39; \ i.e. \ 273+77=350

So, \ there \ are \ 39s 7 \ and \ one \ 77 \ in \ 350

39+1 = 40.

The answer is C.

jinxed · Answer

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

cicerone · 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

PareshGmat · Answer

\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

shrouded1 · Answer

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.

nonameee · Answer

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

deepakh88 · Answer

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 !

enigma123 · Answer

Thanks Bunuel for both the solutions. Very much appreciated.

egiles · Answer

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.

Bunuel · Answer

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.

PerfectScores · Answer

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.

darkhorse19 · Answer

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

russ9 · Answer

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!

Bunuel · Answer

Do we have 175's or 350's to sum? We have 7's and 77's. Try to sum those to get the sum with units digit of 0, and you'll see that the number of terms must be 10, 20, 30, ....

Senthil1981 · Answer

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

GMATinsight · Answer

Check the solution in attachment

CrushGMAT · Answer

There are a few different approaches.

A video explanation of two of them can be found here:
https://www.youtube.com/watch?v=Vcl1CXco3jk

One approach is to think about the maximum and minimum number of 7s we would need in order to reach a sum of 350. The maximum is 350/7 = 50, but 50 isn't an answer choice. We need a number that's lower than 50, so lets find the second-highest number by subtracting a 77 from 350

350 - 77 = 273

273/7 = 39, i.e. our list of numbers could have a 77 and thirty-nine "7s" and the sum would be 350

39 is the correct answer choice. Answer B

The video covers this approach and a second one, as well.

GMATGuruNY · Answer

Let x = the number of 7's and y = the number of 77's.

Total number of terms: 
Since the OA represents the total number of terms, we get:
 x + y = OA .

Sum of the terms: 
Since the sum of the terms is 350, we get:
7x + 77y = 350
7(x + 11y) = 350
 x + 11y = 50 .

Subtracting the red equation from the blue equation, we get:
(x + 11y) - (x + y) = 350 - OA
10y = 350 - OA
OA = 350 - 10y = (multiple of 10) - (multiple of 10) = multiple of 10.

C .

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

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