lucky again with the sum of number

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lucky again with the sum of number

by lukaswelker » Tue Apr 08, 2014 10:32 am
I sadly guessed this question right, but I'm totaly lost in understanding how to properly solve it.

Here's the question

If each term in the sum a1 + a2...+ an is either 7 or 77 and the sum equals 350, which of the following could be n?

38; 39; 40; 41; 42

Any help would be wonderful
Many thanks
Lukas
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by Brent@GMATPrepNow » Tue Apr 08, 2014 10:34 am
If each term in sum a1+a2+ ... +an is either 7 or 77 and the sum equals 350, which of the following could be equal to n?

A)37
B)39
C)40
D)41
E)42
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 thirty-nine 7's and one 77 in the sum, for a total of 40 terms.

This matches one of the answer choices, so the correct answer is C

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by Brent@GMATPrepNow » Tue Apr 08, 2014 10:35 am
If each term in sum a1+a2+ ... +an is either 7 or 77 and the sum equals 350, which of the following could be equal to n?

A)37
B)39
C)40
D)41
E)42
Another possible approach it 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 GMATGuruNY » Tue Apr 08, 2014 11:07 am
If each term in the sum a1+a1+a3+a4+...+aN is either 7 or 77 and the sum equals 350, which of the following could be equal to n?

1) 38
2) 39
3) 40
4) 41
5) 42
Approach 1 (similar to Brent's):

350/7 = 50.
So if each term were 7, there would be 50 terms.
The answer choices are all a little less than 50.
Implication:
Most -- but not all -- of the terms will be 7.

Case 1: 1 term = 77
Amount remaining = 350-77 = 273.
273/7 = 39
This works!
Thus, 1 term = 77, while 39 terms = 7, for a total of 40 terms.

The correct answer is C.

Alternate approach:

Let x = the number of 7's and y = the number of 77's.
Since the sum is 350, we get:
7x + 77y = 350
x + 11y = 50.

The answer choices imply that the total number of terms -- x+y -- is between 38 and 42, inclusive.
Thus:
38 ≤ x+y ≤ 42.
Substituting x = 50-11y into 38 ≤ x+y ≤ 42, we get:
38 ≤ (50-11y) +y ≤ 42
38 ≤ 50 - 10y ≤ 42
-12 ≤ - 10y ≤ -8
12 ≥ 10y ≥ 8.

Only one integer value satisfies the resulting inequality:
y = 1.
Since y=1 and x=50-11y, we get:
x = 50 - 11*1 = 39.
Thus, the total number of terms = x+y = 39+1 = 40.
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by [email protected] » Tue Apr 08, 2014 4:07 pm
Hi lukaswelker,

Both Brent and Mitch have showcased a tactic that is called "brute force." The idea is that certain GMAT Quant questions are based on a limited number of possibilities. Rather than try to come up with a long-winded or complex math approach, sometimes it's best to just map out the possibilities and do the necessary math to prove what the correct answer is. Be on the lookout for these situations because they will occur a number of times on Test Day.

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