If a, b, and c are consecutive integers and 0 < a < b < c, is the prod
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23 Aug 2018, 22:36
Hi,
Another way to solve this question is trying out numbers.
Always try out few numbers and understand the pattern of the answers.
Here in the question, it’s given, a, b and c are consecutive integers and also positive (0 < a < b < c),
Question: Is product abc a multiple of 8 ?
To be a multiple of 8 , it has to have minimum of three 2’s because of 8 can be written in prime factorization as 2^3
Now, lets try out some numbers,
a, b and c minimum has to be 1,2 and 3. Here the product abc is 6. Not a multiple of 8.
If a, b and c is 2,3 and 4. Then the product abc is 24. Multiple of 8.
Okay, then lets check the next set.
If a,b and c is 3,4 and 5. Then the product abc is 60. Not a multiple of 8.
If a, b and c is 4,5 and 6. Then the product abc is 120. Multiple of 8.
Note 1:
We can note from the above pattern, everytime if a and c is even, then the product abc is a multiple of 8(Infact 24).
Note 2:
Also, if a and c are odd, then the product abc need not be a multiple of 8 but it is an even number.
Statement I is sufficient.
Given, the product ac is even.
Since, a, b and c are consecutive integers. If ac is even, then both a and c has to be even.
So, from the note 1, we can see that product is a multiple of 8.
So sufficient.
Statement II is insufficient.
Given, the product bc is a multiple of 4.
bc is a multiple of 4 means, either b is a multiple of 4 and c is odd or b is odd and c is a multiple of 4.
For example,
If a,b and c is 3,4 and 5. Then the product abc is 60. Not a multiple of 8. But here 4*5 = 20 is a multiple of 4.
If a, b and c is 2,3 and 4. Then the product abc is 24. Multiple of 8. Here 3*4= 12 also a multiple of 4.
So it is insufficient.
So the answer is A.
I alone sufficient.