Tricky Tricky Tricky.
As is often my tactic on difficult DS problems, I first approached this by plugging in numbers.
(1) When x is divided by 2y, the remainder is 4.
If x = 10 and 2y = 6, r = 4. Thus, x = 10, y = 3 and r of x/y = 1
If x = 12 and 2y = 8, r = 4. Thus, x = 12, y = 4. and r of x/y = 0
Since I found two different remainders for x/y, I eliminated AD on my AD/BCE grid.
(2) When x+y is divided by y, the remainder is 4.
If y = 5, the first integer that gives a remainder of 4 when divided by y is 9. Thus, x = 4 (since 5+4 = 9) and the remainder of x/y = 4. Keeping y = 5, the next integer that gives a remainder of 4 when divided by y is 14. Thus, x = 9. Since this simply made our original x increase by y (we added one more value of y), the remainder of x/y will remain unchanged (r still = 4). This pattern holds for all y = 5.
If y = 6, the first integer that gives a remainder of 4 when divided by y is 10. Thus, x = 4 and the remainder of x/y = 4. Keeping y = 6, the next integer that gives a remainder of 4 when divided by y is 16. Thus, x = 10. Since this simply made our original x increase by y, the remainder of x/y will remain unchanged (r still = 4). This pattern holds for all y = 6.
If y = 7, the first integer that gives a remainder of 4 when divided by y is 11. Thus, x = 4 and the remainder of x/y = 4. Keeping y = 7, the next integer that gives a remainder of 4 when divided by y is 18. Thus, x = 11. Since this simply made our original x increase by y, the remainder of x/y will remain unchanged (r still = 4). This pattern holds for all y = 7.
In essence, for the remainder the fraction from statement (2) to be 4, the difference between the numerator and the denominator (or a multiple of the demonimator) must be 4. Since, in equation (2), x is the number that accounts for this difference, the first value for x = 4 (with y>4). Thus the remaider of the first value for x/y will always be 4. To keep the remainder of 4 in (2), the numerator will always increase by multiples y. Thus, the remainder of x/y will always = 4. The correct answer is B. (There must be an algebraic method for this problem, but it is too late right now for me to see it. I or another instructor will reply with this during daylight hours!)
-dan
If x and y are positive integers, what is the remainder when x is divided by y?
1) When x is divided by 2y, the remainder is 4.
2) When x+y is divided by y, the remainder is 4.
What do you think the answer is? I think answer is "E", but as per book it is B.
I don't understand why? This example is taken from Kaplan 800 book.