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Let a, b, and c be digits with a ≠ 0$. The three-digit integer abc lie

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Let a, b, and c be digits with a ≠ 0$. The three-digit integer abc lie  [#permalink]

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New post 26 Mar 2019, 23:31
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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

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Re: Let a, b, and c be digits with a ≠ 0$. The three-digit integer abc lie  [#permalink]

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New post 27 Mar 2019, 02:18
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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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Re: Let a, b, and c be digits with a ≠ 0$. The three-digit integer abc lie  [#permalink]

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New post 29 Mar 2019, 07:44
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


The value of the integer abc is 100a + 10b + c and that of acb is 100a + 10c + b. Since abc is ⅓ of the way between two consecutive perfect squares (say n^2 and (n + 1)^2) and acb is ⅔ of the way between the same two perfect squares, the distance (or difference) between abc and acb is ⅓ of the way between the two perfect squares. So we have:

(100a + 10c + b) - (100a + 10b + c) = ⅓[(n + 1)^2 - n^2]

10c + b - 10b - c = ⅓[n^2 + 2n + 1 - n^2]

9c - 9b = ⅓(2n + 1)

27c - 27b = 2n + 1

27(c - b) = 2n + 1

We see that 2n + 1 must be a multiple of 27; therefore, n can be 13, 40, 67, etc. However, n can’t be 40 (or more), since n^2 = 40^2 = 1600 and there are no 3-digit numbers that are between 40^2 and 41^2. So n must be 13. In that case, we have:

27(c - b) = 27

c - b = 1

Since n = 13, n^2 = 169 and (n + 1)^2 = 14^2 = 196. So we can see that a must be 1. So acb must be 187 and abc must be 178. In any case, a + b + c = 1 + 7 + 8 = 16.

Answer: C
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Re: Let a, b, and c be digits with a ≠ 0$. The three-digit integer abc lie   [#permalink] 29 Mar 2019, 07:44
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