# 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? $\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$

# 2018 AMC 8 Problem 21

Number Theory
Easy
##### Suggested Book
Excursion in Mathematics

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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