mehaksal wrote:The three-digit positive integer n can be written as ABC, in which A, B, and C stand for the unknown digits of n. What is the remainder when n is divided by 37?
(1) A+ B/10 + C/100= B+ C/10 + A/100
(2) A + B/10 + C/100= C + A/10 + B/100
suggest some strategy for quickly solving!!
The three-digit integer ABC = 100A + 10B + C.
Put each statement in terms of ONE VARIABLE.
Statement 1: A + B/10 + C/100 = B + C/10 + A/100
Clear the fractions:
100(A + B/10 + C/100) = 100(B + C/10 + A/100)
100A + 10B + C = 100B + 10C + A.
Since the hundreds digit on each side must be the same:
100A = 100B
A=B.
Since the units digit on each side must be the same:
C=A.
Since A=B, we in turn know that C=B.
Thus:
100A + 10B + C = 100B +10B + B = B(100+10+1) = 111B.
Since 111 = 3*37, 111B is a multiple of 37.
Thus, when 111B is divided by 37, the remainder is 0.
SUFFICIENT.
Statement 1: A + B/10 + C/100 = C + A/10 + B/100
Clear the fractions:
100(A + B/10 + C/100) = 100(C/100 + A/10 + A/100)
100A + 10B + C = 100C + 10A + B.
Since the hundreds digit on each side must be the same:
100A = 100C
A=C.
Since the tens digit on each side must be the same:
10B = 10A
B=A.
Since A=C, we in turn know that B=C.
Thus:
100A + 10B + C = 100C +10C + C = C(100+10+1) = 111C.
Since 111 = 3*37, 111C is a multiple of 37.
Thus, when 111C is divided by 37, the remainder is 0.
SUFFICIENT.
The correct answer is
D.
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