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Re: If a + b+ c are integers, is abc divisible by 4? [#permalink]

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26 Jan 2013, 12:28

alexpavlos wrote:

If a + b+ c are integers, is abc divisible by 4?

1) a + b + 2c is even 2) a + 2b + c is odd

Can anyone please show me what is the festet and most "elegant" way of solving this? Do you just try each scenario?

Thanks! Alex

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.

Re: If a + b+ c are integers, is abc divisible by 4? [#permalink]

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26 Jan 2013, 13:04

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.

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!

alexpavlos wrote:

If a + b+ c are integers, is abc divisible by 4?

1) a + b + 2c is even 2) a + 2b + c is odd

Can anyone please show me what is the festet and most "elegant" way of solving this? Do you just try each scenario?

Can anyone please show me what is the festet and most "elegant" way of solving this? Do you just try each scenario?

Thanks! Alex

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.

Re: If a + b+ c are integers, is abc divisible by 4? [#permalink]

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01 Feb 2015, 04:51

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.

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}.
_________________

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?

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.
_________________

Re: If a + b+ c are integers, is abc divisible by 4? [#permalink]

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16 Nov 2016, 11:50

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