A quick lesson on remainders:
When x is divided by 5, the remainder is 3.
In other words, x is 3 more than a multiple of 5:
x = 5a + 3.
When x is divided by 7, the remainder is 4.
In other words, x is 4 more than a multiple of 7:
x = 7b + 4.
Combined, the statements above imply that when x is divided by both 5 and 7 -- in other words, when x is divided by 35 -- there will be a constant remainder R.
Put another way, x is R more than a multiple of 35:
x = 35c + R.
To determine the value of R:
Make a list of values that satisfy the first statement:
When x is divided by 5, the remainder is 3.
x = 5a + 3 = 3, 8, 13, 18...
Make a list of values that satisfy the second statement:
When x is divided by 7, the remainder is 4.
x = 7b + 4 = 4, 11, 18...
The value of R is the SMALLEST VALUE COMMON TO BOTH LISTS:
R = 18.
Putting it all together:
x = 35c + 18.
Another example:
When x is divided by 3, the remainder is 1.
x = 3a + 1 = 1, 4, 7, 10, 13...
When x is divided by 11, the remainder is 2.
x = 11b + 2 = 2, 13...
Thus, when x is divided by both 3 and 11 -- in other words, when x is divided by 33 -- the remainder will be 13 (the smallest value common to both lists).
x = 33c + 13 = 13, 46, 79...
Onto the problem at hand:
gander123 wrote:Got another one:
"What is the remainder when positive integer x is divided by 6?
(1) When x is divided by 2, the remainder is 1; and when x is divided by 3, the remainder is 0.
(2) When x is divided by 12, the remainder is 3."
Correct answer: D
Statement 1: When x is divided by 2, the remainder is 1, and when x is divided by 3, the remainder is 0.
List 1: x = 2a + 1 = 1,
3, 5, 7...
List 2: x = 3b =
3, 6, 9, 12...
The smallest value common to both lists is 3.
Thus:
x = 6c + 3 = 3, 9, 15, 21...
When any value here is divided by 6, the remainder in each case will be 3.
SUFFICIENT.
Statement 2: When x is divided by 12, the remainder is 3.
x = 12a + 3 = 3, 15, 27, 39...
When any value here is divided by 6, the remainder in each case will be 3.
SUFFICIENT.
The correct answer is
D.
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