Question

A number when divided by 7 leaves a remainder 4 and when divided by 9 leaves a remainder 5. What is the smallest such number greater than 100?

Hint:

For remainder problems, check if adding or subtracting a small number from N results in a common multiple. Here, N+3 gives remainder 0 with 7 (\(N=7k+4 \implies N+3=7k+7\)) but not with 9 (\(N=9m+5 \implies N+3=9m+8\)). If it works for both, the solution is faster.

Answer 1

Correct Answer: 158

Step 1: Let the number be \( N=7k+4 \). We need \( 7k+4\equiv5\pmod{9} \), i.e. \( 7k\equiv1\pmod9 \).
Step 2: Since \( 7\times4=28\equiv1\pmod9 \), the inverse of \( 7 \) mod \( 9 \) is \( 4 \), so \( k\equiv4\pmod9 \), meaning \( k=9m+4 \).
Step 3: Then \( N=7(9m+4)+4=63m+32 \), so \( N \) cycles through \( 32, 95, 158, 221,\ldots \) as \( m=0,1,2,3,\ldots \)
Step 4: The first value exceeding 100 is when \( m=2 \).
\[ \boxed{63(2)+32=158} \]