If a, b, and c are consecutive integers, such that a*b*c = a + b + c

Question

If a, b, and c are consecutive integers (a < b < c), such that a*b*c = a + b + c, then how many triplets of (a, b, c) are possible?

A. 0
B. 1
C. 2
D. 3
E. More than 3

dimri10 · Answer

I came out with D.
see no other option than: 1-2-3, negative 1-2-3 and -1,0,1

now if i take (x-1)*x* (x+1)=x-1+x+x+1 i come with x^3-4x
=0
x(x^2-4)=0 or x(x+2)(x-2), so x can be either 0, 2 or -2

RaghavKhanna · Answer

The three integers are of the form: a, a+1, a+2 (since they are consecutive)

=> a(a+1)(a+2) = a + a+1 + a+2
=> a(a+1)(a+2) = 3(a+1)
=> a(a+1)(a+2) - 3(a+1) = 0
=> (a+1)  =0

=> a+1 = 0
=> a = -1

=> a^2 + 2a -3 = 0
=> a^2 + 3a - a -3 =0
=> a(a+3) - 1(a-3) = 0

=> (a-1)(a+3) = 0
 => a = 1,- 3

When a = 1, b = 2, c = 3
When a = -1, b = 0, c = 1
When a = -3, b = -2, c = -1

i.e; Three possible triplets.

Answer: D

satya2029 · Answer

a(a+1)(a+2)=a+a+1+a+2
a(a+1)(a+2)=3(a+1)
(a+1) =0
a=-1 or a= 1 or a=3
D:)

yashikaaggarwal · Answer

Let three consecutive numbers be (A-1)(A)(A+1)
Therefore, 
(A-1)(A)(A+1) = (A-1)+(A)+(A+1)
(A^2-A)(A+1) = 3A
(A^3-A^2+A^2-A) = 3A
(A^3-A) = 3A
A^3 = 4A
A^2 = 4
A = ±2
So the two consecutive triplets will be (-1,-2,-3) & (1,2,3)
Also,
The minimum value of sum of integers can be = 0
(A-1)(A)(A+1) = 0
(A^2-A)(A+1) = 0
(A^3-A^2+A^2-A) = 0
(A^3-A) = 0
A^3 = A
A^2 = 0
A = 0
The possible consecutive triplet is (-1,0,1)
Total no. Of possible triplets are 3 => (-1,0,1)(1,2,3)(-1,-2,-3)

Answer is D

Posted from my mobile device

Kinshook · Answer

Asked: If a, b, and c are consecutive integers (a < b < c), such that a*b*c = a + b + c, then how many triplets of (a, b, c) are possible?

a = b-1
c = b+1

(b-1)b(b+1) = 3b
b(b^2-1) = 3b; b = {0,-2,2}

(a ,b, c) = {(-1,0,1), (-3,-2,-1),(1,2,3)}

IMO D

ScottTargetTestPrep · Answer

Solution:

Since the product of three positive integers is usually much greater than the sum of the same three integers, and if a*b*c is indeed equal to a + b + c, then the integers must be very small. Upon trial and error, we see that 1*2*3 = 1 + 2 + 3. Since a, b, and c are consecutive integers and a < b < c, we see that (a, b, c) could be (1, 2, 3). If a = 2, we see that 2*3*4 = 24 and 2+3+4 = 9. We see that a can’t be greater than 1. Let’s try values that are less than 1.

If a = 0, 0*1*2 = 0 and 0 + 1 + 2 = 3, we see that a*b*c ≠ a + b + c.

If a = -1, -1*0*1 = 0 and -1 + 0 + 1 = 0, we see that a*b*c = a + b + c.

If a = -2, -2*-1*0 = 0 and -2 + (-1) + 0 = -3, we see that a*b*c ≠ a + b + c.

If a = -3, -3*-2*-1 = -6 and -3 + (-2) + (-1) = -6. we see that a*b*c = a + b + c.

If a = -4, -4*-3*-2 = -24 and -4 + (-3) + (-2) = -9. we see that a*b*c ≠ a + b + c.

We can stop at this point. We see that if a < -4, the product of a, b, and c will be much less than the sum of a, b, and c. Therefore, there are three values for a, namely, 1, -1, and -3, and hence there are three triplets of (a, b, c) that are possible.

Alternate Solution:

We see that b = a + 1 and c = a + 2. Let’s substitute these values in a*b*c = a + b + c:

a(a + 1)(a + 2) = a + (a + 1) + (a + 2)

a(a^2 + 2a + a + 2) = 3a + 3

a^3 + 3a^2 + 2a = 3a + 3

a^3 + 3a^2 - a - 3 = 0

To solve this cubic equation, let’s factor by grouping by factoring a^2 from the first two terms and factoring -1 from the last two terms:

a^2(a + 3) + (-1)(a + 3) = 0

(a + 3)(a^2 - 1) = 0

a + 3 = 0 or a^2 - 1 = 0

The first equation yields a solution of a = -3. The second equation yields the solutions a = 1 or a = -1. We can verify that all these values satisfy a * (a + 1) * (a + 2) = a + (a + 1) + (a + 2). Thus, there are three triplets satisfying the given equation.

Answer: D

BlueRocketAsh · Answer

a(a+1)(a+2)=a+a+1+a+2 
 a(a+1)(a+2)=3a+3 
 a(a+1)(a+2)=3(a+1) 
 a(a+1)(a+2)-3(a+1)=0 
 (a+1) =0 
 (a+1)(a^2+2a-3)=0 
 (a+1)(a-3)(a+1)=0 
 a=-1 or 3 or 1

There are three distinct roots
Therefore there are three different sets.

hr1212 · Answer

Nice question!

Let the numbers be x-1, x and x+1

Then,
a*b*c = a + b + c
 (x-1)*x*(x+1) = 3x  --- Note: Don't cancel out x from both the sides or you will miss solution x=0
 x*(x^2-1) - 3x = 0
x*(x^2 - 4) = 0  
 x = 0  or  x^2-4=0  
x = 0 or x = 2 or x = -2

So the triplets are (-1, 0, 1) , (1, 2, 3) , (-3, -2, -1)

Answer: D

If a, b, and c are consecutive integers, such that abc = a + b + c

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

If a, b, and c are consecutive integers, such that a*b*c = a + b + c

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

If a, b, and c are consecutive integers, such that abc = a + b + c