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:
ela07mjt wrote:When the positive integer n is divided by 25, the remainder is 13. What is the value of n ?
(1) n < 100
(2) When n is divided by 20, the remainder is 3.
According to the question stem:
n = 25a + 13 = 13, 38, 63, 88...
Statement 1: n<100
n could be any of the values in the list above.
INSUFFICIENT.
Statement 2: When n is divided by 20, the remainder is 3.
Thus:
n = 20b + 3 = 3, 23, 43, 63...
When we combine this condition with that in the question stem, we know the following:
n must be a multiple of 20 and 25 -- in other words, a multiple of 100 -- plus the smallest value common to both lists (63).
Thus:
n = 100c + 63 = 63, 163, 263...
Since n can be more than one value, INSUFFICIENT.
Statements combined:
The only value that satisfies both statements is n=63.
SUFFICIENT.
The correct answer is
C.
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