Find the Smallest Number Which Leaves Remainders 8 and 12 When Divided by 28 and 32

Video Explanation

Watch the video below to understand the complete solution step by step:

Solution

Question: Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.

Step 1: Form the Mathematical Conditions

If the required number is N, then:

N − 8 is divisible by 28
N − 12 is divisible by 32

Step 2: Express N in Terms of One Variable

Let N − 8 = 28k
⇒ N = 28k + 8

Substitute this value of N in the second condition:

28k + 8 − 12 is divisible by 32

⇒ 28k − 4 is divisible by 32

⇒ 28k ≡ 4 (mod 32)

⇒ −4k ≡ 4 (mod 32)

⇒ 4k ≡ 28 (mod 32)

⇒ k ≡ 7 (mod 8)

Step 3: Find the Smallest Value of N

Smallest value of k = 7

N = 28 × 7 + 8

N = 196 + 8

N = 204

Final Answer

∴ The smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively is 204.

Conclusion

Thus, by forming equations using the given remainders and solving them, we find that the required smallest number is 204.