AMC 8 2018 Problem 21 Number Theory

Understand the problem

How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11? $\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$
Source of the problem

2018 AMC 8 Problem 21

Topic
Number Theory
Difficulty Level
Easy
Suggested Book
Excursion in Mathematics

Start with hints

Do you really need a hint? Try it first!
If the number is \( N \) , then \( N = 6a +2 = 9b +5 = 11c +7 \) . [where \( a, b \ and \ c \in Z^+ \) ] .
\( N = 6a +2 = 9b +5 = 11c +7 \) . [where \( a, b \ and \ c \in Z^+ \) ] \( \Rightarrow N = 6(a+1)- 4 =9(b+1) -4 = 11(c+1) -4 \)
Now think of a number which is simultaneously multiple of 6 , 9 and 11 i.e. \( lcm (6,9 ,11 ) \). Then our required number \( N \) is just less than 4 of \( lcm (6,9 ,11 ) \).
So, to find out such numbers the least common multiple of these number is \( 11 \times 3^2 \times 2 \), so the numbers that fulfill this can be written as \( 198k -4 \), where $k$ is a positive integer. \( \\ \) This value is only a three digit integer when \(k\) is \( 1, 2 ,3 ,4 \ and \ 5 \), which gives and \( 194 , 392 , 590 , 788, \ and \ 986 \) respectively. Thus we have 5 values . \( \\ \) Note Basically this problem is the direct application of CHINESE REMAINDER THEOREM . But in this solution the advantages of values are taken .

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