If a, b, and c are positive integers, with a < b < c, are a, b, and c

i approached the question differently ... this may not be the best way but i'm going to share it anyway ... sharing is caring

a, b, c > 0 [positive integers]
b=a+1, c=a+2 [a, b, c are consecutive integers]

(1) 1/a - 1/b = 1/c
1/a - 1/a+1 = 1/a+2
1/a(a+1) = 1/(a+2)
a+2 = a(a+1)
c = a*b [c=a+2, b=a+1]

if a, b, c are consecutive ... the above equation in bold can never be true
2 ? 0*1
3 ? 1*2
4 ? 2*3

1 is sufficient to explain that a, b, c are not consecutive integers

(2) a+c = (b-1)(b+1)
a+c = a*c [a=b-1, c=b+1]

if a, b, c are consecutive ... the above equation in bold can never be true
0+2 ? 0*2
1+3 ? 1*3
2+4 ? 2*4

2 is sufficient to explain that a, b, c are not consecutive integers

ans is D

Hey guys - The way I did it was

(1) 1/a – 1/b = 1/c ---> 1/a = 1/b + 1/c

LCM of 'b' and 'c' would never be smaller than b or c because these numbers are co-prime which means that they do not share any common integer factor other than 1.

Now since c>b>a then they following the logic above they can't be consecutive integers. Sufficient

(2) a + c = b^2 – 1 ---> a+c = (b+1)(b-1)

Now if a,b,c were consecutive integers it will be true that (a+c)/2 is equal to b
So we would have that a+c = 2b

Now if we replace a+c = 2b in the first equation we would have that

2b = b^2-1

And when trying to solve this quadratic we would realize that b is not an integer, since we would have to use the formula
So it contradicts the information that a,b,c are integers. Hence a,b,c cannot be consecutive integers

Insuff

Answer (D)

Please let me know whether this makes sense.
If it does, throw me some Kudos!

Cheers
J

Hi Bunuel,

Thanks for the answer. One question - what is the meaning of "this eq has no positive integer root ?". It is because the eq has A = +- 2 ?

We are given that \(a\) is a positive integer, while from \(a^2=2\), \(a=-\sqrt{2}\) or \(a=\sqrt{2}\). Hence our assumption that a, b, and c were consecutive integers was wrong.

Not sure if this was the correct approach or not...

statement 1: sufficient
If we say a=1 b=2 c=3 then this equation would be 1/1 - 1/2 = 1/3 which is not true. Also seems to be right for all the other quick consecutive #s I plugged in. I'm going to say that this is sufficient by showing us that they are not consecutive.

statement 2: sufficient
If we say a=1 b=2 c=3 then this equation would be 1+3=2^2 - 1
4=4-1
4=3
That also does not work so I'm going to say that this is sufficient for the same reason as above.

Answer D!

I tried the problem in the following way:

let a = n-1, b = n and c = n+1, assuming that the numbers are consecutive.

from statement 1-
1/(n-1) - 1/n = 1/(n+1); this reduces to n^2-2n-1 = 0. This equation does not have any integer roots. i.e. n = 1+sqrt(2) or n = 1- sqrt(2). so consecutive numbers do not satisfy this equation. sufficient to prove that a,b and c are not consecutive.

from statement -2:
n + 1 = n^2 - n, this also reduces to n^2-2n-1 = 0. same reasoning as above. sufficient.

therefore answer is : [D]

testing both the statements with simple consecutive no.1,2 & 3

statement 1: 1/a -1/b =1/c
Clearly it shows a,b &c are not consecutive.


statement 2: a+c=b^2-1
This statement too gives the definite result that the no.s are not consecutive.


Answer is D

Official Solution:

If \(a\), \(b\), and \(c\) are positive integers, with \(a \lt b \lt c\), are \(a\), \(b\), and \(c\) consecutive integers?

The question can be rephrased "Is \(b = a + 1\) and is \(c = a + 2\)?"

One way to approach the statements is to substitute these expressions involving \(a\) and solve for \(a\). Since this could involve a lot of algebra at the start, we can just substitute \(a + 1\) for \(b\) and test whether \(c = a + 2\), given that both are integers.

Statement 1: SUFFICIENT.

Following the latter method, we have
\(\frac{1}{a} - \frac{1}{a + 1} = \frac{1}{c}\)
\(\frac{a + 1}{a(a + 1)} - \frac{a}{a(a + 1)} = \frac{1}{c}\)
\(\frac{1}{a(a + 1)} = \frac{1}{c}\)
\(a^2 + a = c\)

Now we substitute a + 2 for c and examine the results:
\(a^2 + a = a + 2\)
\(a^2 = 2\)

\(a\) is the square root of 2. However, since \(a\) is supposed to be an integer, we know that our assumptions were false, and \(a\), \(b\), and \(c\) cannot be consecutive integers.

We can now answer the question with a definitive "No," making this statement sufficient.

We could also test numbers. Making \(a\) and \(b\) consecutive positive integers, we can solve the original equation \((\frac{1}{a} - \frac{1}{b} = \frac{1}{c})\). The first 4 possibilities are as follows:
\(\frac{1}{1} - \frac{1}{2} = \frac{1}{2}\)
\(\frac{1}{2} - \frac{1}{3} = \frac{1}{6}\)
\(\frac{1}{3} - \frac{1}{4} = \frac{1}{12}\)
\(\frac{1}{4} - \frac{1}{5} = \frac{1}{20}\)

Examining the denominators, we can see that \(c = ab\). None of these triples so far are consecutive, and as \(a\) and \(b\) get larger, \(c\) will become more and more distant, leading us to conclude that \(a\), \(b\), and \(c\) are not consecutive.

Statement 2: SUFFICIENT.

Let's try substituting \((a + 1)\) for \(b\) and \((a + 2)\) for \(c\).
\(a + a + 2 = (a + 1)^2 - 1\)
\(2a + 2 = a^2 + 2a\)
\(2 = a^2\)

Again, we get that \(a\) must be the square root of 2. However, we know that \(a\) is an integer, so the assumptions must be false. We can answer the question with a definitive "No," and so the statement is sufficient.

Answer: D.

The question says that a, b & c are positive integers.
To verify if they are consecutive, we can substitute 'a+1' in place of b and 'a+2' in place of c in each of the equations.

(1) Solution a=sq rt. of 2. As the value of a as per (1) is not an integer, a, b and c cannot be consecutive integers.
(2) Solution a=sq rt. of 2. As the value of a as per (2) is not an integer, a, b and c cannot be consecutive integers.

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