Bunuel wrote:
Let a, b, and c be digits with a ≠ 0$. The three-digit integer abc lies one third of the way from the square of a positive integer to the square of the next larger integer. The integer acb lies two thirds of the way between the same two squares. What is a+b+c?
(A) 10
(B) 13
(C) 16
(D) 18
(E) 21
As per the question ,
(100a+10c+b) - (100a + 10b + c) = 9(c-b)
From the below diagram we can see that 9(c-b) is 1/3 of the difference between the two consecutive squares.
i.e \(((x+1)^2 - x^2)/3\) = 9(c-b)
=> \(((x+1)^2 - x^2)\)= 27(c-b)...(1)
Now is x is odd then x+1 will be even ,so will be their squares. Similarly if x is even then x+1 will be odd.
So sum of their squares will odd+even or even+odd i.e a odd number
For 27(c-b) to be odd . c-b = 1,3,5,6
For c-b =1 Equation 1 becomes \(((x+1)^2 - x^2)\)=27 => 2x + 1 = 27 => x = 13
So the two squares will be 13^2 and 14^2 i.e 169 and 196
For c-b =3 Equation 1 becomes \(((x+1)^2 - x^2)\)=81 => 2x + 1 = 81 => x = 40
So the two squares will be 40^2 and 41^2 i.e 1600 and 1681 . But here the numbers are greater than 3 digits.
So c-b=1 is only possible i.e x=13
so integer abc =13^2 + (14^2 - 13^2)/3 = 169+ (196 - 169)/3 = 169 +27/3 = 178
So a+b+c = 1 + 7 + 8 = 16. Answer (C).
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