lunarpower wrote:courtesy of user 'karenmeow'
if each term in sum a1+a2+...+an is either 7 or 77, and the sum equals 350, which can be n?
choices: 38, 39, 40, 41, 42
Solution:
We can let x be the number of terms that are 7 and y be the number of terms that are 77. Notice that x + y = n, the total number of terms. We can also set up the following equation
7x + 77y = 350
Dividing both sides by 7 we have:
x + 11y = 50
11y = 50 - x
y = (50 - x)/11
Because y is an integer, we know that (50 - x)
must be a multiple of 11.
Thus, to make that true, variable x can only be 6, 17, 28 or 39.
If x = 6, y = 4, so n = 10.
If x = 17, y = 3, so n = 20.
If x = 28, y = 2, so n = 30.
If x = 39, y = 1, so n = 40.
We see that answer choice C is the only one that matches our possibilities for n.
Note: In looking at the equation
x + 11y = 50, we also could have taken a units digit approach. We know that the units digit of x + 11y is a 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 x + 11y will result in the
same units digit as x + y. Since we know that x + 11y = 50, we know that x + 11y has a units digit of zero. This means that x + y (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.
The answer is
C
