If each term in the sum a1 + a2 + ... aN is either a 7 or 77 and the sum equals 350, which of the following could be equal to N?
38
39
40
41
42
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What is N?
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srcc25anu
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Lets try with ONE 77
350 - 77 = 273 and 273 = 39 SEVENs
N = 1 + 39 = 40
Ans C
Also lets try for TWO 77s
350 - 154 = 196 and 196 = 28 SEVENs
N = 2 + 28 = 30 (not an answer choice)
As we go on increasing the # of 77s, N will continue to decrease which is not any of the available answer choices.
Only choice that satisfies this is N = 40
350 - 77 = 273 and 273 = 39 SEVENs
N = 1 + 39 = 40
Ans C
Also lets try for TWO 77s
350 - 154 = 196 and 196 = 28 SEVENs
N = 2 + 28 = 30 (not an answer choice)
As we go on increasing the # of 77s, N will continue to decrease which is not any of the available answer choices.
Only choice that satisfies this is N = 40
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aaggar7
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The unit's digit is 7 (as 7 and 77 both have the same unit digit) and the sum (350 ) has the unit digit as 350.So,the only way we can get unit's digit as 0 ,is by multiplying the sum with a multiple of 10.
so C
so C
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GMATGuruNY
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Approach 1: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
7*50 = 350.
Thus, if each term is 7, the total number of terms = 50.
The answer choices are all JUST A BIT LESS than 50.
Thus, MOST of the terms must be 7.
If 1 term = 77, the amount remaining = 350-77 = 273.
273/7 = 39.
This works: 1 term = 77, 39 terms = 7.
Total number of terms = 1+39 = 40.
The correct answer is C.
Approach 2:
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.
Of the answer choices, only 40 is a multiple of 10.
The correct answer is C.
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Brent@GMATPrepNow
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Notice that 77 does not divide into 350 many times.Rudy414 wrote:If each term in the sum a1 + a2 + ... aN is either a 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
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
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Brent@GMATPrepNow
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Another possible approach is to look for a pattern.Rudy414 wrote:If each term in the sum a1 + a2 + ... aN is either a 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
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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