Bunuel wrote:
Which of the following pairs of numbers (x,y) is NOT a valid solution to the equation 8933x − 5920y = 97?
A. (4749, 7166)
B. (10669, 16099)
C. (16588, 25031)
D. (22509, 33965)
E. (28429, 42898)
This looked like a monster problem. Then I saw the key: 5920. Its units digit is 0.
Thus the second term (5920*y) in every answer will end in 0:
(---0 * anything) = last digit of 0
y doesn't matter.
The second term can be written
(---0)So: (----
3)*x) - (----0) = 7
The first number, 8933, ends in 3.
It will be multiplied by x.
And the product of (3*x)
must end in 7because 97 ends with 7: _7
We subtract the second number's last digit of 0 from the last digit of (3*x).
(Product of 3*x) - (----0) = _7
(---
7) - (----0) = _7
So check only the first term in the options.
(3 * x) must = _7
If (3 * x) must end in 7, then
the units digit of x must be 9.
(3*1)=3. (3*2)=6 ... Only (3*9)=2
7 works.
(---3)*
(9) = ----7
All the answers except (C) end in 9.
Check C if you want to be sure
If x were (C), it would yield (
16588 *
8933) = nnnn
4 \((4 - 0)\neq{7}\)
(first term) - (second term) cannot = 97
Not a valid solution.
Answer C