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If each term in the sum \(a_1+a_2+a_3+\cdots+a_n\) is either

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by Brent@GMATPrepNow » Tue Oct 01, 2019 11:29 am
AAPL wrote:GMAT Prep

If each term in the sum \(a_1+a_2+a_3+ \cdots +a_n\) 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

OA 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

Cheers,
Brent
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by Brent@GMATPrepNow » Tue Oct 01, 2019 11:30 am
AAPL wrote:GMAT Prep

If each term in the sum \(a_1+a_2+a_3+ \cdots +a_n\) 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

OA 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
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by Scott@TargetTestPrep » Mon Oct 07, 2019 6:52 pm
AAPL wrote:GMAT Prep

If each term in the sum \(a_1+a_2+a_3+ \cdots +a_n\) 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

OA C
We can let a be the number of terms that are 7 and b be the numbers that are 77. Notice that a + b = n, the total number of terms. Now we can set up the following equation

7a + 77b = 350

Dividing both sides by 7 we have:

a + 11b = 50

11b = 50 - a

b = (50 - a)/11

So we see that (50 - a) must be a multiple of 11.

Thus, to make that true, variable a can be only 6, 17, 28, or 39.

If a = 6, b = 4, so n = 10.

If a = 17, b = 3, so n = 20.

If a = 28, b = 2, so n = 30.

If a = 39, b = 1, so n = 40.

We see that answer C is the only one that matches our possibilities for n.

Alternate Solution:

In looking at the equation a + 11b = 50, we can take a units digit approach. We know that the units digit of a + 11b is zero. We also know that 11 times any number will result in the same units digit as the original number. As an example let's consider the number 2. We see that 2 has a units digit of 2, and 2 x 11 = 22 also has a units digit of 2.

Thus, we can say that a + 11b will result in the same units digit as a + b. Since we know that a + 11b = 50, we know that a + 11b has a units digit of zero. This means that a + b (or "n" in this case) also has a units digit of zero. Since 40 is the only answer choice with a units digit of zero we know that it is the correct answer.

Answer: C

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