Three consecutive positive integers are raised to the first, second an

Question

Three consecutive positive integers are raised to the first, second and third powers respectively and then added. The sum so obtained is a perfect square whose square root equals the total of the three original integers. What is the value of the smallest of the three integers ?

A. 1
B. 2
C. 3
D. 4
E. 5

Are You Up For the Challenge: 700 Level Questions

A. 1
B. 2
C. 3
D. 4
E. 5

Are You Up For the Challenge: 700 Level Questions

GMATinsight · Answer

Rather than making algebraic equation, I would like to try with a few examples

(1, 2, 3)
 1^1+2^2+3^3 = 1+4+27 = 32  Not a perfect square so OUT

(2, 3, 4)
 2^1+3^2+4^3 = 2+9+64 = 75  Not a perfect square so OUT

(3, 4, 5)
 3^1+4^2+5^3 = 3+16+125 = 144  a perfect square

√144 = 12 = (3+4+5) All conditions met

Smallest = 3

Answer: Option C

Archit3110 · Answer

trick to this question to see on given answer options
so the consective integers can be 1,2,3 ; 2,3,4; 3,4,5 ; 4,5,6 ; 5,6,7
start from middle value
3,4,5 ; given condition raise power for each digit ; 3+16+125 ; 144 ; sum of digits ; 12 whose square value is 144
so sufficient
IMO C: 3

raunakd11 · Answer

3 Consecutive integers are 3, 4 and 5. These numbers are given the power of 1, 2 and 3 respectively, 
3^1+4^2+5^3= 3+16+125
 =144
Square root of 144 is 12, which is the sum of 3, 4 and 5.
The smallest number out of it is 3.
Which is Option C

Cheers!

Regards,
Raunak Damle

nick1816 · Answer

3 consecutive numbers are 'a-2', 'a-1' and 'a'

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

a^3+a^2-a-1=9(a-1)^2

(a^3-1) + (a^2-a) - 9(a-1)^2 = 0 
 
 (a-1)(a^2+a+1) + a(a-1) - 9(a-1)^2 = 0

(a-1) (a^2+a+1+a-9a+9) =0

(a-1) (a^2-7a+10) = 0

(a-1)(a-2)(a-5)=0

As all 3 integers are positive, a is equal to 5.

smallest of 3 integer = 5-2=3

KarishmaB · Answer

Let the three original numbers be (n - 1), n and (n+1)

Their sum = (n - 1) + n + (n + 1) = 3n

Then the perfect square will be 9n^2. This is the sum of (n - 1), n^2 and (n + 1)^3

(n - 1) + n^2 + (n + 1)^3 = 9n^2

n^3 - 5n^2 + 4n = 0

n(n^2 - 5n + 4) = 0

n = 0, 1, 4
Then (n - 1) can be -1, 0 or 3

Answer (C)

lacktutor · Answer

a^{1}+ (a+1)^{2} + (a+2)^{3}= (a+ a+1 +a+2)^{2}

-->  a+ a^{2}+ 2a +1 + a^{3}+ 6a^{2}+ 12a +8 = (3a+3)^{2}= 9a^{2}+ 18a+ 9 
 a^{3} -2a^{2} -3a =0

a*(a +1)*(a -3) =0 
 a=0 ,  a=-1  and  a=3 
--> a is positive integer -->  a=3

The answer is C

ScottTargetTestPrep · Answer

Solution:

We can let x = the largest positive integer, and so that the two smaller ones are x - 2 andx - 1, respectively. We can create the equation:

x - 2 + (x - 1)^2 + x^3 = (x - 2 + x - 1 + x)^2

x - 2 + x^2 - 2x + 1 + x^3 = (3x - 3)^2

x^3 + x^2 - x - 1 = 9x^2 - 18x + 9

x^3 - 8x^2 + 17x - 10 = 0

x^3 - 8x^2 + 7x + 10x - 10 = 0

x(x^2 - 8 + 7) + 10(x - 1) = 0

x(x - 1)(x - 7) + 10(x - 1) = 0

(x - 1)(x^2 - 7x + 10) = 0

(x - 1)(x - 2)(x - 5) = 0

x = 1 or x = 2 or x = 5

If the largest integer x = 1, then the smallest integer, which is (x - 2), would not be positive. The same holds if x = 2 because x - 2 would equal 0. Therefore, x must be 5, and hence the smallest of the three integers is 3.

Alternate Solution:

For those who have difficulties in factoring a cubic equation, we can solve the problem by verifying the given choices (especially when they are such small numbers).

A. 1

If the smallest integer is 1, we have: 1 + 2^2 + 3^3 = 1 + 4 + 27 = 32. However, since 32 is not a perfect square (as required), A is not the correct answer.

B. 2

If the smallest integer is 2, we have: 2 + 3^2 + 4^3 = 2 + 9 + 64 = 75. However, since 75 is not a perfect square (as required), B is not the correct answer.

C. 3

If the smallest integer is 3, we have: 3 + 4^2 + 5^3 = 3 + 16 + 125 = 144. Since 144 = 12^2 and 12 is the sum of 3, 4, and 5, C is the correct answer.

Answer: C

sambitspm · Answer

See the attachment

bumpbot · Answer

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

Three consecutive positive integers are raised to the first, second an

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

Three consecutive positive integers are raised to the first, second an

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