luisnavarro wrote:
Bunuel wrote:
Official Solution:
(1) \(2ab\) is divisible by 5. If \(a=b=5\) then the answer is YES but if \(a=5\) and \(b=1\) then the answer is NO. Not sufficient.
(2) \(a-b\) is divisible by 5. If \(a=b=5\) then the answer is YES but if \(a=b=1\) then the answer is NO. Not sufficient.
(1)+(2) From (2) \(a-b\) is divisible by 5 so \((a-b)^2=(a^2+b^2)-2ab\) is also divisible by 5. Next, since from (1) \(2ab\) is divisible by 5 then \(a^2+b^2\) must also be divisible by 5 in order their sum to be divisible by 5. Sufficient.
Answer: C
Hi Bunuel,
Could you explain statement (2)...... Why If a=b=5 then the answer is YES? (5) -(5) = 0..... Does cero is a divisor of 5?
Thanks a lot.
Regards.
Luis Navarro
Looking for 700
0 is not a divisor of any integer, but a multiple of every integer. So, 0 is divisible by 5.
ZERO:1. 0 is an integer.
2. 0 is an even integer. An even number is an
integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.
3. 0 is neither positive nor negative integer (the only one of this kind).
4. 0 is divisible by EVERY integer except 0 itself.Check more here:
number-properties-tips-and-hints-174996.htmlSo Cero is multiple of every integer (That is why is divisible by any integer except cero) but not a factor or divisor of any integer.
Thanks a lot Bunuel...