Latest Elitmus Aptitude Question SOLUTION: Three positive consecutive integers are raised to the first, second and third power respectively and then added. The sum so obtained is perfect square whose square root is equal to

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(#M40037976) ELITMUS Category question Keep an EYE

Three positive consecutive integers are raised to the first, second and third power respectively and then added. The sum so obtained is perfect square whose square root is equal to the total of the three original integers. Which of the following best describes the minimum, say m, of these three integers?
A. 1≤m≤3
B.4≤m≤6
C.7≤m≤9
D.10≤m≤12
E.3≤m≤15

Asked In Elitmus Liti (9 years ago)

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Prateek Pandey  (9 years ago)

option E.

from the options given
only one triplet is obtaining this property
3,4,5
3^1+4^2+5^3=144
sqrt(144)=12=3+4+5

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Mani kandan  (9 years ago)

the smallest of these integers is m. The other two are m+1 and m+2 because it is given that the three integers are consecutive positive integers.
A number raised to the first power is the number itself.
A number raised to the second power is squared or multiplied by itself.
A number raised to the third power is cubed or multiplied by itself and then multiplied by itself again.
So we have,
m + (m+1)*(m+1) + (m+2)*(m+2)*(m+2)
The square root of this is equal to
m+m+1+m+2

We now have the equation
m+m+1+m+2 = sq rt [m + (m+1)*(m+1) + (m+2)*(m+2)*(m+2)]
Solve this quadratic equation to get
m=0, m=3, m= -1

m=-1 is not possible because the question says it should be a positive integer, so choice 1 describes the answer best.

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pranav khandelwal  (9 years ago)

OPTION C
AS 8+9^2+10^3=1089 WHOSE SQRT=33 WHICH IS NOT EQUAL TO 27

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