Deepthi Subbu wrote:If each term in the sequence a1+a2+...... +an is either a 7 or 77 , and the sum equals 350.,which of the following could be the value of n?

1.38
2.39
3.40
4.41
5.42

OA 40

350/7 = 50. So if each term were 7, we'd have 50 terms. The answer choices are all a little less than 50. So most -- but not all -- of the terms will be 7.

If 1 term = 77, we have 350-77 = 273 left.
273/7 = 39. This works!
So 1 term = 77, 39 terms = 7.
Total number of terms is 1+39 = 40.

The correct answer is C.

Hi,

This is from GMATPrep 1, right?

a(1) to a(n) can be either 7 or 77. Whatever they are, they always have unit digit of "7".

Note that the sum of them is 350. Which means the number of "a" must be a multiple of 10. Let's take this example:

7+ xx7 + x7+ 7+ xxx7+ .... + 7 = 35350, how many number are there in the series? Choices are 10,11,12,13 and 14. It must be 10.
The sum of "n" numbers ended in "7" will have the last digit of:

n = 1, Last digit (LD) = 7
n = 2, LD = 4
n = 3, LD= 1
n = 4, LD = 8
n = 5, LD = 5
n=6, LD = 2
n= 7, LD = 9
n=8, LD = 6
n=9, LD = 3
n =10, LD =0
n=11, LD = 7
......

Go back to case, only answer choice C: 40 is a multiple of 10. Thus C must be the correct answer.
The trick here is to see the digit number of the answer choices.

To go a little further, if they ask us to give what value "n" could be, then the answer is:
n = 10, with 4*77 + 6*7 ( four 77, six 7)
n = 20, with 3*77 + 17*7
n= 30, with 2*77 + 28*7
n=40, with 1*77 + 39*7 (this is the case of the question)
n =50, with 0*77 + 50*7

Hope this helps.

Deepthi Subbu wrote:If each term in the sequence a1+a2+...... +an is either a 7 or 77 , and the sum equals 350.,which of the following could be the value of n?

1.38
2.39
3.40
4.41
5.42

Let m of the n parcels be 7´s (0<= m <=n) ; the other (n-m) parcels will certainly be 77´s.

We have m7 + (n-m)77 = 350 and doing trivial calculations you get 11n = 10m+50 = 10(m+5).

From the fact that m is an integer, 11n is a multiple of 10. From the fact that 10 and 11 are relatively prime, that means that n (also an integer) must be divisible by 10. We are done.

Regards,
Fabio.

GMATGuruNY wrote:

Deepthi Subbu wrote:If each term in the sequence a1+a2+...... +an is either a 7 or 77 , and the sum equals 350.,which of the following could be the value of n?

1.38
2.39
3.40
4.41
5.42

OA 40

350/7 = 50. So if each term were 7, we'd have 50 terms. The answer choices are all a little less than 50. So most -- but not all -- of the terms will be 7.

If 1 term = 77, we have 350-77 = 273 left.
273/7 = 39. This works!
So 1 term = 77, 39 terms = 7.
Total number of terms is 1+39 = 40.

The correct answer is C.

I made a very ugly mistake , calculated 350-77 and marked the answer as 39. I dint realize where was I going wrong until u added the 1(77) to 39 . Thanks GMATGuruNY.

limestone wrote:Hi,

This is from GMATPrep 1, right?

a(1) to a(n) can be either 7 or 77. Whatever they are, they always have unit digit of "7".

Note that the sum of them is 350. Which means the number of "a" must be a multiple of 10. Let's take this example:

7+ xx7 + x7+ 7+ xxx7+ .... + 7 = 35350, how many number are there in the series? Choices are 10,11,12,13 and 14. It must be 10.
The sum of "n" numbers ended in "7" will have the last digit of:

n = 1, Last digit (LD) = 7
n = 2, LD = 4
n = 3, LD= 1
n = 4, LD = 8
n = 5, LD = 5
n=6, LD = 2
n= 7, LD = 9
n=8, LD = 6
n=9, LD = 3
n =10, LD =0
n=11, LD = 7
......

Go back to case, only answer choice C: 40 is a multiple of 10. Thus C must be the correct answer.
The trick here is to see the digit number of the answer choices.

To go a little further, if they ask us to give what value "n" could be, then the answer is:
n = 10, with 4*77 + 6*7 ( four 77, six 7)
n = 20, with 3*77 + 17*7
n= 30, with 2*77 + 28*7
n=40, with 1*77 + 39*7 (this is the case of the question)
n =50, with 0*77 + 50*7

Hope this helps.

You are correct , this is from GMATPrep1. Great explanation , thank you!