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What is the sum of all integer solutions to 1 < (x - 2)^2 < 25 ? 22 Apr 2019, 01:44
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B
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Difficulty:

75% (hard)

Question Stats:

57% (02:10) correct 43% (02:03) wrong based on 333 sessions

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What is the sum of all integer solutions to 1 < (x - 2)^2 < 25 ?

(A) 10
(B) 12
(C) 15
(D) 19
(E) 25
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B
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What is the sum of all integer solutions to 1 < (x - 2)^2 < 25 ? 28 Apr 2019, 21:13
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Bunuel
What is the sum of all integer solutions to 1 < (x - 2)^2 < 25 ?

(A) 10
(B) 12
(C) 15
(D) 19
(E) 25
Show more

\(1\) < \((x-2)^2\) < \(25\)
\(\sqrt{1}\) < \(| x-2 |\) < \(\sqrt{25}\)

Solving \(1 < x-2 < 5\), \(3 < x < 7\).
\(x = 4, 5, 6\)

Solving \(1 < -(x-2) < 5\), \(-1 > x-2 > -5\), \(1 > x > -3\).
\(x = 0, -1, -2\)

Sum of all integer solutions: 4+5+6+0+(-1)+(-2) = 12
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What is the sum of all integer solutions to 1 < (x - 2)^2 < 25 ? Updated on: 22 Apr 2019, 03:29
Originally posted by LevanKhukhunashvili on 22 Apr 2019, 02:53.
Last edited by LevanKhukhunashvili on 22 Apr 2019, 03:29, edited 1 time in total.
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Learn from the mistakes of others, find whats wrong with ranges below:

Lets take range for
\((x-2)^2\)>1

\(x^2\)+4-2x-1>0
(x-2)(x-1)>0
Range: -inf<x<1; 2<x<+ inf

Range for
\((x-5)^2\)<25
-3<x<7

Combining both ranges we have got -3<x<1; 2<x<7
Possible INTEGRAL values are: -2, -1, 0, 3, 4, 5, 6

Their sum= 15

IMO
Ans: C
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What is the sum of all integer solutions to 1 < (x - 2)^2 < 25 ? Updated on: 22 Apr 2019, 03:15
Originally posted by Archit3110 on 22 Apr 2019, 02:59.
Last edited by Archit3110 on 22 Apr 2019, 03:15, edited 1 time in total.
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Bunuel
What is the sum of all integer solutions to 1 < (x - 2)^2 < 25 ?

(A) 10
(B) 12
(C) 15
(D) 19
(E) 25
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1 < (x - 2)^2 < 25
valid for x=-2,-1,0,4,5,6
sum = 12
IMO B
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What is the sum of all integer solutions to 1 < (x - 2)^2 < 25 ? 22 Apr 2019, 03:11
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1 < (x - 2)^2 < 25
valid for x= -2,-1,0,4,5,6
sum = 12
IMO
Ans:B

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What is the sum of all integer solutions to 1 < (x - 2)^2 < 25 ? 22 Apr 2019, 03:24
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you must exclude 3, as (x - 2)^2 is equal to 1 when x=3
LevanKhukhunashvili
Lets take range for
\((x-2)^2\)>1

\(x^2\)+4-2x-1>0
(x-2)(x-1)>0
Range: -inf<x<1; 2<x<+ inf

Range for
\((x-5)^2\)<25
-3<x<7

Combining both ranges we have got -3<x<1; 2<x<7
Possible INTEGRAL values are: -2, -1, 0, 3, 4, 5, 6

Their sum= 15

IMO
Ans: C
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What is the sum of all integer solutions to 1 < (x - 2)^2 < 25 ? 28 Apr 2019, 21:34
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(x-2)^2 value between 1 and 25 can be 4, 9, 16 and x-2 can be either positive or negative values of 2, 3 and 4.
Equating these positive/negative values will give sum 12

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What is the sum of all integer solutions to 1 < (x - 2)^2 < 25 ? 26 Jan 2020, 13:28
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Bunuel
What is the sum of all integer solutions to 1 < (x - 2)^2 < 25 ?

(A) 10
(B) 12
(C) 15
(D) 19
(E) 25
Show more

\(square:1 < (x - 2)^2 < 25…1 < |x - 2| < 5\)
\(positive:1 < x - 2 < 5…3<x<7…x=[4,5,6]\)
\(negative:1 < -(x-2) < 5…1<-x+2<5…-1<-x<3…-3<x<1=[-2,-1,0]\)
\(sum.solutions:[4,5,6]+[-2,-1,0]=15+(-3)=12\)

Ans (B)
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What is the sum of all integer solutions to 1 < (x - 2)^2 < 25 ? 22 Jun 2024, 09:27
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What is the sum of all integer solutions to 1 < (x - 2)^2 < 25 ?

(A) 10
(B) 12
(C) 15
(D) 19
(E) 25

1 < (x - 2)^2 < 25
Therefore (x - 2) can be negative or positive.
Since integer solutions are required the squares of those integers are only 4,9,16

Thus,  (x - 2)^2 = 4 giving x = 4(when x-2>0 i.e. 2^2), 0(when x-2<0 i.e. (-2)^2)
(x - 2)^2 = 9 giving x = 5(when x-2>0), -1(when x-2<0)
(x - 2)^2 = 16 giving x = 6(when x-2>0), -2(when x-2<0)
Sum = 4+0+5-1+6-2 = 12

Answer B.
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What is the sum of all integer solutions to 1 < (x - 2)^2 < 25 ? 19 Oct 2025, 18:34
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