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2. Find the greatest number of 5 digits, that will give us remainder of 5, when divided by 8 and 9
respectively.
(a) 99921
(b) 99931
(c) 99941
(d) 99951


Sagot :

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