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 [tex]\(N\)[/tex] such that when [tex]\(N\)[/tex] is divided by both 8 and 9, it leaves a remainder of 5. Mathematically, this can be expressed as:
[tex]\[
N \equiv 5 \pmod{8}
\][/tex]
and
[tex]\[
N \equiv 5 \pmod{9}
\][/tex]
2. Mathematical formulation: If [tex]\(N \equiv 5 \pmod{8}\)[/tex], this means that:
[tex]\[
N = 8k + 5 \text{ for some integer } k.
\][/tex]
Similarly, if [tex]\(N \equiv 5 \pmod{9}\)[/tex], this means that:
[tex]\[
N = 9m + 5 \text{ for some integer } m.
\][/tex]
3. Combining the two conditions:
Since both expressions for [tex]\(N\)[/tex] 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 [tex]\(N\)[/tex] must satisfy:
[tex]\[
N \equiv 5 \pmod{72}
\][/tex]
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 [tex]\(N\)[/tex] that is less than or equal to 99999 and satisfies [tex]\(N \equiv 5 \pmod{72}\)[/tex].
5. Verification and calculation:
[tex]\[
99999 \div 72 \approx 1388.875
\][/tex]
Hence, the highest multiple of 72 less than 99999 is:
[tex]\[
72 \times 1388 = 99936
\][/tex]
Adding the remainder 5:
[tex]\[
N = 99936 + 5 = 99941
\][/tex]
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
