If each term in the sum a1+a2+a3+.....+an 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
The OA is C.
I'm really confused with this PS question. Experts, any suggestion about how to solve it? Thanks in advance.
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If each term in the sum a1+a2+a3+.....+an is either 7 or...
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APPROACH #1:LUANDATO wrote:If each term in the sum a1+a2+a3+.....+an 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
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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APPROACH #2:LUANDATO wrote:If each term in the sum a1+a2+a3+.....+an 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
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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We only need to regard last digit of everything!
The last digit of both 07 and 77 is 7.
Now multiply this 7 by the last digit of each given answer:
A. 8 x 7 = 56
B. 9 x 7 = 63
C. 0 x 7 = 0
D. 1 x 7 = 7
E. 2 x 7 = 14
Only answer C has a last digit that matches the last digit of 350. None of the other answers are possible.
The last digit of both 07 and 77 is 7.
Now multiply this 7 by the last digit of each given answer:
A. 8 x 7 = 56
B. 9 x 7 = 63
C. 0 x 7 = 0
D. 1 x 7 = 7
E. 2 x 7 = 14
Only answer C has a last digit that matches the last digit of 350. None of the other answers are possible.
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Let x = the number of 7's and y = the number of 77's.LUANDATO wrote:If each term in the sum a1+a2+a3+.....+an 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
Total number of terms:
Since the OA represents the total number of terms, we get:
x + y = OA.
Sum of the terms:
Since the sum of the terms is 350, we get:
7x + 77y = 350
7(x + 11y) = 350
x + 11y = 50.
Subtracting the red equation from the blue equation, we get:
(x + 11y) - (x + y) = 350 - OA
10y = 350 - OA
OA = 350 - 10y = (multiple of 10) - (multiple of 10) = multiple of 10.
The correct answer is C.
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Let m = number of multiples of 77
and p = number of multiples of 7
Note that 77 = 70 + 7, so 77m = 70m + 7m
Therefore 350 = 70m + 7m + 7p = 70m + 7(m+p)
As 350 is a multiple of 10, then (m+p) must be a multiple of 10 too
Only Answer C gives n = 40 = m+p = multiple of 10
and p = number of multiples of 7
Note that 77 = 70 + 7, so 77m = 70m + 7m
Therefore 350 = 70m + 7m + 7p = 70m + 7(m+p)
As 350 is a multiple of 10, then (m+p) must be a multiple of 10 too
Only Answer C gives n = 40 = m+p = multiple of 10
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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 equationLUANDATO wrote:If each term in the sum a1+a2+a3+.....+an 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
7a + 77b = 350
Divided both sides by 7 we have:
a + 11b = 50
11b = 50 - a
b = (50 - a)/11
So we know that 50 - a must be a multiple of 11.
Thus, to make that true, variable a can only be 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.
Note: In looking at the equation a + 11b = 50, we also could have taken 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 take the number 2. 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 is the correct answer.
Answer: C
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