Find the smallest number which when divided by 24 and 36 leaves a remainder of 11
The correct answer and explanation is:
To find the smallest number that leaves a remainder of 11 when divided by 24 and 36, we need to solve a system of congruences. Specifically, the number must satisfy the following conditions:
- x≡11 (mod 24)x \equiv 11 \ (\text{mod} \ 24)
- x≡11 (mod 36)x \equiv 11 \ (\text{mod} \ 36)
This means that the number xx when divided by both 24 and 36 should leave a remainder of 11 in both cases.
Step 1: Express the number in terms of a general solution
For the first congruence x≡11 (mod 24)x \equiv 11 \ (\text{mod} \ 24), the number xx can be written as: x=24k+11x = 24k + 11
where kk is an integer.
For the second congruence x≡11 (mod 36)x \equiv 11 \ (\text{mod} \ 36), substitute x=24k+11x = 24k + 11 into this equation: 24k+11≡11 (mod 36)24k + 11 \equiv 11 \ (\text{mod} \ 36)
Simplifying this: 24k≡0 (mod 36)24k \equiv 0 \ (\text{mod} \ 36)
This means that 24k24k must be divisible by 36.
Step 2: Solve for kk
Since 24k24k must be divisible by 36, divide both sides of the equation by 12: 2k≡0 (mod 3)2k \equiv 0 \ (\text{mod} \ 3)
This means that kk must be divisible by 3. So, let k=3mk = 3m, where mm is an integer.
Step 3: Substitute k=3mk = 3m back into the equation for xx
Substitute k=3mk = 3m into x=24k+11x = 24k + 11: x=24(3m)+11=72m+11x = 24(3m) + 11 = 72m + 11
Thus, the general solution for xx is: x=72m+11x = 72m + 11
Step 4: Find the smallest solution
To find the smallest number, set m=0m = 0 (since we want the smallest value): x=72(0)+11=11x = 72(0) + 11 = 11
Thus, the smallest number that satisfies both conditions is x=11x = 11.
Conclusion
The smallest number that when divided by 24 and 36 leaves a remainder of 11 is 11.