Let's denote the required number by \( N \). We know that \( N \) leaves the same remainder of 9 when divided by 15, 18, and 36.
So, \( N = 15a + 9 = 18b + 9 = 36c + 9 \), where \( a, b, \) and \( c \) are integers.
Now, let's find the least common multiple (LCM) of 15, 18, and 36, which is the smallest number that is divisible by all three:
\[ \text{LCM}(15, 18, 36) = 180 \]
So, \( N \) is of the form \( 180k + 9 \), where \( k \) is an integer.
Now, we need to find the smallest value of \( k \) such that \( N \) is divisible by 11. For \( N = 180k + 9 \) to be divisible by 11, \( 180k \) must have the same remainder as \( -9 \) when divided by 11.
\[ 180k \equiv -9 \pmod{11} \]
Now, find the modular inverse of 180 (mod 11) to simplify:
\[ 180 \cdot 5 \equiv 1 \pmod{11} \]
Multiply both sides by 5:
\[ k \equiv -9 \cdot 5 \equiv 7 \pmod{11} \]
So, \( k = 7 \) is the smallest value that satisfies the condition.
Now, substitute \( k = 7 \) into the expression for \( N \):
\[ N = 180 \cdot 7 + 9 = 1269 \]
Finally, find the sum of the digits of \( N \):
\[ \text{Sum of digits} = 1 + 2 + 6 + 9 = 18 \]
Therefore, the correct answer is A) 18.
shortcut method
1. Find the least common multiple (LCM) of 15, 18, and 36, which is 180.
2. The required number is of the form \(180k + 9\), where \(k\) is an integer.
3. For the number to be divisible by 11, we need to find the smallest \(k\) such that \(180k\) leaves the same remainder as \(-9\) when divided by 11.
4. Calculate \(180 \cdot 5 \equiv 1 \pmod{11}\) to find the modular inverse of 180 (mod 11).
5. Multiply \(-9\) by the modular inverse \(5\) to find \(k \equiv 7 \pmod{11}\).
6. Substitute \(k = 7\) into the expression for the number to find \(N = 180 \cdot 7 + 9 = 1269\).
7. Finally, sum the digits of \(N\) to get \(1 + 2 + 6 + 9 = 18\).
So, the sum of the digits of the least number is 18.
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