"I take a number that leaves a remainder of 8 when divided by 12. I take another number which is 4 short of a multiple of 12. Their sum will always be a multiple of 8", claims Snehal. Examine his claim and justify your conclusion.
Given: First number: leaves remainder 8 when divided by 12 → form: 12k + 8 Second number: 4 short of a multiple of 12 → form: 12m - 4
Claim: Their sum will always be a multiple of 8.
Analysis: Sum = 12k + 8 + 12m - 4 = 12(k + m) + 4 = 4[3(k + m) + 1]
For divisibility by 8: need 3(k + m) + 1 to be even This requires k + m to be odd
Counterexample: k = 0, m = 0: Sum = 4 (not divisible by 8)
Supporting example: k = 0, m = 1: Sum = 16 (divisible by 8)
Conclusion: Snehal's claim is sometimes true, not always. The sum is divisible by 8 only when k + m is odd.
Explanation
The chapter discusses divisibility properties. For the sum to always be a multiple of 8, it should work for all values of k and m. But 12(k + m) + 4 is divisible by 8 only when 3(k + m) + 1 is even. Example: k = 0, m = 0 gives sum = 4 (not divisible by 8). But k = 0, m = 1 gives sum = 16 (divisible by 8). So the claim is sometimes true.
Solution Steps
First number has form: 12k + 8 (leaves remainder 8 when divided by 12)
Second number has form: 12m - 4 (4 short of a multiple of 12)
Sum = 12k + 8 + 12m - 4 = 12(k + m) + 4
Factor: 12(k + m) + 4 = 4[3(k + m) + 1]
For divisibility by 8: need 3(k + m) + 1 to be even
This requires k + m to be odd
Counterexample: k = m = 0 gives sum = 4 (not divisible by 8)
Conclusion: Claim is sometimes true, not always true