a, b, and c are three distinct positive integers. What is the product abc?
1) a + b + c = 7
2) ab + bc + ca = 14
I understand that statement one is sufficient but i can't work my way out in statement 2.
pls help
three distinct positive integers
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Assume that 0<a<b<c
For the 1st statement:
If a >= 2, we have a+b+c >= 2+3+4=9 > 7 => incorrect. So a must equal 1.
To continue with b. b>a means b>1.
If b >= 3, we have a+b+c >= 1+3+4=8 > 7 => incorrect. So b must equal 2.
Then, we find c=7-1-2=4. Statement 1 is sufficient.
Similarly, we apply this way for 2nd statement and find (a,b,c) = (1,2,4)
Therefore, each one is sufficient.
For the 1st statement:
If a >= 2, we have a+b+c >= 2+3+4=9 > 7 => incorrect. So a must equal 1.
To continue with b. b>a means b>1.
If b >= 3, we have a+b+c >= 1+3+4=8 > 7 => incorrect. So b must equal 2.
Then, we find c=7-1-2=4. Statement 1 is sufficient.
Similarly, we apply this way for 2nd statement and find (a,b,c) = (1,2,4)
Therefore, each one is sufficient.
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For Statement 1, if we make all of the integers greater than 1, the smallest that they can be is 2, 3 and 4.a, b, and c are three distinct positive integers. What is the product abc?
1) a + b + c = 7
2) ab + bc + ca = 14
2 + 3 + 4 = 9. That's already too big. So one of them has to be 1.
1 + 3 + 4 = 8 Too big.
Must be 1 + 2 + 4. So there is only one set that works, and Statement 1 is sufficient.
For Statement 2, if we make all of the integers greater than 1, the smallest that they can be is 2, 3, and 4.
Plugging in 2, 3 and 4, we get 6 + 12 + 8 = 26 Too big. So one must be 1.
If a is 1 then the equation becomes b + bc + c = 14
Hmm. I'm thinking 8 + 6. That works with b = 2 and c = 4. 2 + 8 + 4 = 14.
Can other numbers work? If b = 3 and c = 4, the sum is going to be greater than 14.
So the integers must be 1, 2 and 4. So Statement 2 is sufficient.
The correct answer is D.
Marty Murray
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Contact me at [email protected] for a free consultation.