Discussion
Rajesh had to arrange his books in uniform groups. He makes groups of 4 books each. But 3 books are left. He tries it with groups of 5 books each. But still 3 books are left. 3 books are still left when he tried with groups of 9 or 10 books each. How many books does he have?
Answer: D
Three books are left over in every arrangement, so the total is 3 more than a common multiple of all four group sizes:
\[ N = \operatorname{LCM}(4,5,9,10)\,k + 3 \]
Taking prime factors, \( 4 = 2^{2} \), \( 5 = 5 \), \( 9 = 3^{2} \), \( 10 = 2 \cdot 5 \):
\[ \operatorname{LCM} = 2^{2} \cdot 3^{2} \cdot 5 = 180 \]
\[ \therefore\; N = 180k + 3, \quad\text{smallest such } N = 183 \]
The other options are all multiples of the LCM with nothing left over, so each would divide exactly and leave no remainder at all.
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