x is a positive integer less than 500. When x is divided by 7, the remainder is 1; when x is divided by 3, the remainder is 2. How many x are there?
A. 21
B. 22
C. 23
D. 24
E. 25
A. 21
B. 22
C. 23
D. 24
E. 25
Onto the problem at hand: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 35 -- the LOWEST COMMON MULTIPLE OF 5 AND 7 -- 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 33 -- the LCM of 3 and 11 -- the remainder will be 13 (the smallest value common to both lists).
x = 33c + 13 = 13, 46, 79...
When x is divided by 7, the remainder is 1.ziyuenlau wrote:x is a positive integer less than 500. When x is divided by 7, the remainder is 1; when x is divided by 3, the remainder is 2. How many x are there?
A. 21
B. 22
C. 23
D. 24
E. 25
x = 7a + 1 = 1, 8...
When x is divided by 3, the remainder is 2.
x = 3b + 2 = 2, 5, 8...
Thus, when x is divided by 21 -- the LCM of 7 and 3 -- the remainder will be 8 (the smallest value common to both lists).
x = 21c + 8.
Since x must be less than 500, we get:
21c + 8 < 500
21c < 492
c < 23.4.
Since c can be any integer between 0 and 23, inclusive -- for a total of 24 values -- there are 24 options for x.
The correct answer is D.
