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by Brent@GMATPrepNow » Fri May 10, 2013 6:34 am
kkadvent wrote:If each term in a1, a2, a3,.....,an is either 7 or 77 and the sum is 350 what is the value of n?

a)39
b)40
c)41
d)42
e)43
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 B

Cheers,
Brent
Last edited by Brent@GMATPrepNow on Sun May 26, 2013 4:40 am, edited 1 time in total.
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by Brent@GMATPrepNow » Fri May 10, 2013 6:38 am
kkadvent wrote:If each term in a1, a2, a3,.....,an is either 7 or 77 and the sum is 350 what is the value of n?

a)39
b)40
c)41
d)42
e)43
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 B is a multiple of 10, this must be the correct answer.

Cheers,
Brent
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by sairakarim07 » Fri May 10, 2013 11:28 am
kkadvent wrote:If each term in a1, a2, a3,.....,an is either 7 or 77 and the sum is 350 what is the value of n?

a)39
b)40
c)41
d)42
e)43
The important part of the problem is that sum is 350 which shows 0 as a unit digit.
Summation of the sequence will give 0 (zero) as unit digit when there are 10 (multiple of 10) 7's/77's. That means 7 or 77 times 10 or multiple 10 (20,30,40 and so on) produces 0 as unit digit.
For this reason The answear is B
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by sairakarim07 » Fri May 10, 2013 11:36 am
Brent@GMATPrepNow wrote:
kkadvent wrote:If each term in a1, a2, a3,.....,an is either 7 or 77 and the sum is 350 what is the value of n?

a)39
b)40
c)41
d)42
e)43
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 B is a multiple of 10, this must be the correct answer.

Cheers,
Brent

thanks for your elaborate explanation.
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by GMATGuruNY » Fri May 10, 2013 11:53 am
kkadvent wrote:If each term in a1, a2, a3,.....,an is either 7 or 77 and the sum is 350 what is the value of n?

a)39
b)40
c)41
d)42
e)43
Let x = the number of 7's and y = the number of 77's.
Since the sum of all of the terms is 350, we get:
7x + 77y = 350
7(x + 11y) = 350
x + 11y = 50.

Since a(n) is the last term in the sequence, the total number of terms = n.
Thus:
x+y = n.

Subtracting the second equation from the first, we get:
(x + 11y) - (x+y) = 50-n
10y = 50-n
n = 50-10y
n = 10(5-y).
The resulting equation implies that the value of n must be a multiple of 10.

The correct answer is B.
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