If m and n are positive integers, is the remainder of (10^m + n)/3 larger than the remainder of (10^n + m)/3 ?
1. m > n
2. The remainder of n/3 is 2
When a POWER OF 10 is divided by 3, the remainder will always be 1.
10/3 = 3 R1.
100/3 = 33 R1.
1000/3 = 333 R1.
From there, the remainders will proceed in the following CYCLE: 2, 0, 1...2, 0, 1...2, 0, 1...
To illustrate:
(10 + 1)/3 = 11/3 = 3 R2.
(10 + 2)/3 = 12/3 = 4 R0
(10 + 3)/3 = 13/3 = 4 R1.
(10 + 4)/3 = 14/3 = 4 R2.
(10 + 5)/3 = 15/3 = 5 R0.
(10 + 6)/3 = 16/3 = 5 R1.
(100 + 1)/3 = 101/3 = 33 R2.
(100 + 2)/3 = 102/3 = 34 R0
(100 + 3)/3 = 103/3 = 34 R1.
(100 + 4)/3 = 104/3 = 34 R2.
(100 + 5)/3 = 105/3 = 35 R0.
(100 + 6)/3 = 106/3 = 35 R1.
Notice that the cycle of remainders is the SAME whether integer values are added to 10 or to 100.
Thus, the question stem can be rephrased in terms of 10:
Is the remainder of (10 + n)/3 larger than the remainder of (10 + m)/3?
Statement 1: m>n
Case 1: m=5 and n=2
(10 + n)/3 = 12/3 = 4 R0.
(10 + m)/3 = 15/3 = 5 R0.
Here, the remainders are EQUAL, so the answer to the question stem is NO,
Case 2: m=5 and n=1
(10 + n)/3 = 11/3 = 3 R2.
(10 + m)/3 = 15/3 = 5 R0.
Here, the first remainder of (10 + n)/3 is GREATER, so the answer to the question stem is YES.
INSUFFICIENT.
Statement 2: The remainder of n/3 is 2
Options for n are 5, 8, 11, 14...
If n=5, then (10 + n)/3 = 15/3 = 5 R0.
If n=8, then (10 + n)/3 = 18/3 = 6 R0.
If n=11, then (10 + 11)/3 = 21/3 = 7 R0.
The examples above illustrate that -- in every case -- the remainder of (10 + n)/3 will be 0.
Thus, the answer to the question stem is NO:
Since the remainder of (10 + n)/3 is 0, it is not possible that the remainder of (10 + n)/3 is larger than the remainder of (10 + m)/3.
SUFFICIENT.
The correct answer is
B.
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It was a good observation, and on a lot of questions it's critical to consider things like that.
