To solve the problem of finding the greatest 5-digit number that gives a remainder of 5 when divided by both 8 and 9:
1. Understand the conditions: We need a number \(\(N\)\) such that when \(\(N\)\) is divided by both 8 and 9, it leaves a remainder of 5. Mathematically, this can be expressed as:
\(\[
N \equiv 5 \pmod{8}
\]\)
and
\(\[
N \equiv 5 \pmod{9}
\]\)
2. Mathematical formulation: If \(\(N \equiv 5 \pmod{8}\)\), this means that:
\(\[
N = 8k + 5 \text{ for some integer } k.
\]\)
Similarly, if \(\(N \equiv 5 \pmod{9}\)\), this means that:
\(\[
N = 9m + 5 \text{ for some integer } m.
\]\)
3. Combining the two conditions:
Since both expressions for \(\(N\)\) must hold true simultaneously, we can use the least common multiple (LCM) of 8 and 9 to form a combined modulus condition. The LCM of 8 and 9 is 72. Therefore, the number \(\(N\)\) must satisfy:
\(\[
N \equiv 5 \pmod{72}
\]\)
4. Finding the largest 5-digit number satisfying the condition:
To find the greatest 5-digit number that satisfies this condition, we look for the largest number \(\(N\)\) that is less than or equal to 99999 and satisfies \(\(N \equiv 5 \pmod{72}\)\).
5. Verification and calculation:
\(\[
99999 \div 72 \approx 1388.875
\]\)
Hence, the highest multiple of 72 less than 99999 is:
\(\[
72 \times 1388 = 99936
\]\)
Adding the remainder 5:
\(\[
N = 99936 + 5 = 99941
\]\)
Therefore, the greatest 5-digit number that leaves a remainder of 5 when divided by both 8 and 9 is 99941.
The correct answer is:
(c) 99941
