Naina1 wrote:
D it would be..
1) A divided by 4 leaves a remainder of 1 and A+B divided by 4 leaves a remainder of 3. So, B should leave a remainder of 2.
Eg : A=5 A+B=11 => B=6
B/4 ->remainder 2
This is true for all numbers.
Sufficient
2) A*B divided by 4->remainder is 2
In this case remainders will be multiplied. A leaves remainder of 1, so B should leave a remainder of 2.
Eg : A=9 A*B=18 => B=2
remainder is 2
True for all numbers.
Sufficient
A little hint about how to decide this task quickly and without picking of numbers.
When we have task about division different numbers on the same divisor we can sum or multiple the equations. For example
\(7 = 5*X + 2\)
\(11 = 5*Y + 1\)
if we sum this equations we received such result
\(7+11 = 5*Z + (2+1)\) so 18 / 5 give us remainder 3
of if we multiple this equations we received:
\(7*11 = 5*R + 2\) so 77 / 5 give us remainder 2
In our task we can write this phrase "When integer A divide by 4 remainder is 1, when integer B divided by 4 the remainder is X." in such view:
\(A = 4s + 1\)
\(B = 4r + X\)
and when we sum this equations we received
\(A+B = 4z + (1 + X)\)
and from first statement we know that "When A + B divided by 4 remainder is 3"
So we can make infer that \(1 + X = 3\) and \(X = 2\)
and when we multiple this equations we received
\(A*B = 4z + (1 * X)\)
and from second statement we know that "When A * B divided by 4 remainder is 2"
So we can make infer that \(1 * X = 2\) and \(X = 2\)
This approach can look a little convoluted, but if you try some examples with not variables but numbers, you'll catch the idea and will save a lot of time on such tasks about remainders with similar divisors.
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