Subtract both of then b - c is odd ....one of b or c is odd and other even
Add 2a + 3b + 3c = odd ....one of b or c is odd and other even
Combine...
Let b = odd
c = even
a = odd
Satisfies both premises and Ans for main statement Yes

Let b = even c = odd a = even
Again satisfies both but Ans for main statement No.


So Ans E

Note that
\(O + E =O\), \(E + E = E\), and \(O + O= E\)

To prove whether abc is divisible by 4 when
two of the numbers even (divisible by 2)
One of the numbers is divisible by 4

1) a + b + 2c is even
INSUFFICIENT:
a*b*c not divisible by 4-> Each a, b and c can be ODD (i.e. a*b*c is ODD), but 'a + b + 2c = O+O+2*O=E' will be EVEN.
a*b*c divisible by 4-> If a, b and c are EVEN, 'a + b + 2c' will be even.

2) a + 2b + c is odd
INSUFFICIENT:
a*b*c not divisible by 4-> consider a & b as ODD and c as even (only divisible by 2), 'a + 2b + c' will be ODD and a*b*c will be EVEN, but will be only divisible by 2 (not 4).
a*b*c is EVEN & divisible by 4-> consider a & b as EVEN and c as ODD, 'a + 2b + c' will be ODD, but a*b*c will be divisible by 4.

Combining (1) and (2)
INSUFFIENT: Adding the statements gives 2a + 3b + 3c = ODD, which tells b & c are ODD. However it doesn't tell whether 'a' is even.
If a is even then 2a is divisible by 4. If a is ODD then 2a is NOT divisible by 4.

Hence choice(E) is the answer.
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The easiest way to solve this problem would require the below knowledge

Odd+Odd=Even
Even+Even=Even
Odd+Even=Odd
Even+Odd=Odd

Now the first question just states that a,b,c are integers. We still do not know whether they are even or odd. To be divisible by 4, one of the below 2 scenarios need to occur

1. Either 2 numbers are even
2. One number is a multiple of 4

Now let's take the given statements one by one

1. This tells us that

a+b+2c is even. Since, 2c is always even irrespective of whether c is even or odd, we have no information about c. However, since 2c is even, we know that a+b also needs to be even for the sum to be even. This gives rise to two scenarios

i. a=even, b=even
ii. a=odd and b= odd

Since, we have no further information to determine which of the two is true, we cannot proceed with this. INSUFFICIENT.

2. This tells us that a+2b+c=odd

Since, we know that 2b is always even irrespective of whether b is even or odd, we have no idea about b. However, since 2b is even and the sum is odd, a+c needs to be odd since only a sum of even and odd adds up to an odd number. This can happen only in two ways

i. a=odd and c=even
ii a=even and c=odd

Since, again we have no further info, we cannot proceed further with these statements. INSUFFICIENT.

Together, we still get the four conclusions that we got from 1 and 2.

i. a=even, b=even
ii. a=odd and b= odd
i. a=odd and c=even
ii a=even and c=odd

However, there is no overlap between these four distinct scenarios. Hence, INSUFFICIENT.

Answer=E

This approach might initially confuse you but the more you practice this approach, the lesser time this will take to solve such problems.

Hope it helps!

I am assuming that the question means that a, b and c are integers. This is how I would evaluate the statements:

1) a + b + 2c is even
This means that 'a' and 'b' are either both odd or both even. abc may or may not divisible by 4. Not sufficient.

2) a + 2b + c is odd
This means that one of 'a' and 'c' is odd and the other is even. abc may or may not divisible by 4. Not sufficient.

If 'a' and 'b' both are odd, c must be even. If c is divisible by 4, abc is divisible by 4. Otherwise not. Not sufficient. I needn't even consider the case when 'a' and 'b' are both even.

Answer (E)
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Good question +1. (Note: Although I don't know why you put a+b+c are integers, should be a,b,c are integers)

Anyways, let's nail

1) a+b+2c is Even

This means that a+b must be even

So either a,b, are both even or a,b both odd

Insufficient

2) a+2b+c is odd

So this tells us that either a odd and c even, or the other way around.

Still not sufficient

1+2) Both together we have the following

a+b Even
a+c Odd

Add em up: 2a + b+ c is Odd

Therefore b+c is odd

So either b odd and c even, or the other way around

Let's see first case

b odd, c even, a odd. Not divisible by 4
b even, c odd, a even. Divisible by 4.

Therefore E

Hope its clear
Cheers
J

Can someone please explain how can "abc" be divisible by 4 when either of the below scenarios occurs?
-when two of the numbers are even
-when one of the numbers is divisible by 4

I thought a number is divisible by 4 only if the number formed by the last "two digits" is divisible by 4.

Am I missing something here?

Does those statements contradict each other?

For a number to be divisible by 2 the last digit must be divisible by 2 (so the last digit must be even);
For a number to be divisible by 4 the last two digits must be divisible by 4 (04, 08, 12, 16, ..., 96);
For a number to be divisible by 8 the last three digits must be divisible by 8 (008, 012, 016, ..., );
etc.

If out of two integers, x and y, both are even (x=2m, y=2n), then xy will be a multiple of 4: xy = 2m*2n = 4(mn). For example, 2*6 = 12 = {multiple of 4}.
If out of two integers, x and y, one is a multiple of 4 (for example, if x=4m), then xy will be a multiple of 4: xy = 4m*y = 4(my). For example, 4*5 = 20 = {multiple of 4}.
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a, b and c are not the digits of a three digit number. abc is actually a*b*c where a, b and c are independent numbers.
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all here are assuming that a, b, and c are intergers,but the question does not clarify this.
so why should we have to do so.
what if they are not integers?
please experts help me in this case!