alex1233 wrote:
What is the value of the two-digit positive integer n?
(1) When n is divided by 5, the remainder is equal to the tens digit of n.
(2) When n is divided by 9, the remainder is equal to the tens digit of n.
Algebraic approach:
Any two-digit integer can be represented as 10T+U, where T = the tens digit and U = the units digit.
Thus:
n = 10T+U
Statement 1: When n is divided by 5, the remainder is equal to the tens digit of n.In other words, n is equal to a multiple of 5 plus a remainder of T:
n = 5a+T
Since n = 10T+U and n = 5a+T, we get:
10T+U = 5a+T
U = 5a-9T
Case 1: T=1, implying that U = 5a-9
If a=2, then U=1 --> n = TU = 11
If a=3, then U=6 --> n = TU = 16
Since n can be different values, INSUFFICIENT.
Statement 2: When n is divided by 9, the remainder is equal to the tens digit of n.In other words, n is equal to a multiple of 9 plus a remainder of T:
n = 9b+T
Since n = 10T+U and n = 9b+T, we get:
10T+U = 9b+T
U = 9b-9T
U = 9(b-T)
Since digit U must be a nonnegative integer no greater than 9, the equation above implies that b-T=0 and U=0 or that b-T=1 and U=9.
Case 1: U=0 and b-T=0, implying that T=b
If T=b=1, then n = TU = 10
If T=b=2, then n = TU = 20
Since n can be different values, INSUFFICIENT.
Statements combined:Case 1: U=0
Since U=0 and U = 5a-9T, we get:
0 = 5a-9T
9T = 5a
\(T = \frac{5a}{9}\)
Here, the smallest option for T occurs when a=9, with the result that T=5.
Since remainder T must be less than the divisor of 5 in Statement 1, Case 1 is not viable.
Case 2: U=9
Since U=9 and U = 5a-9T, we get:
9 = 5a-9T
9T = 5a - 9
\(T = \frac{5a}{9} - 1\)
Since remainder T must be less than the divisor of 5 in Statement 1, the equation above implies that a=9 and \(T = \frac{5*9}{9} - 1 = 4\)
Since T=4 and U=9, n = 49
SUFFICIENT.