rxs0005 wrote:
Are i , j , k consecutive integers
The remainder when i + j + k is divided by 3 is 2
The remainder when i*j*k is divided by 3 is 1
I think it's meant that \(i\), \(j\), and \(k\) are integers.
NOTES:Out of any 3 consecutive integers exactly one is divisible by 3. So the product of ANY 3 consecutive integers is divisible by 3 (so remainder is zero).
Consecutive integers can be represented as ... \(n-1\), \(n\), \(n+1\), ... where \(n\) is an integer. Sum of 3 consecutive integers would be \((n-1)+n+(n+1)=3n\), so the sum of ANY 3 consecutive integers is divisible by 3 (so remainder is zero).
More generally:• If \(k\) is odd, the sum of \(k\) consecutive integers is always divisible by \(k\). Given \(\{9,10,11\}\), we have \(k=3\) consecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3.
• If \(k\) is even, the sum of \(k\) consecutive integers is never divisible by \(k\). Given \(\{9,10,11,12\}\), we have \(k=4\) consecutive integers. The sum of 9+10+11+12=42, therefore, is not divisible by 4.
• The product of \(k\) consecutive integers is always divisible by \(k!\), so by \(k\) too. Given \(k=4\) consecutive integers: \(\{3,4,5,6\}\). The product of 3*4*5*6 is 360, which is divisible by 4!=24.
(1) The remainder when i + j + k is divided by 3 is 2 --> as the remainders is not zero, then \(i\), \(j\), and \(k\) are not consecutive integers. Sufficient.
(2) The remainder when i*j*k is divided by 3 is 1--> as the remainders is not zero, then \(i\), \(j\), and \(k\) are not consecutive integers. Sufficient.
Answer: D.
Hope it helps.