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CAT 2025 Slot 2 QA Question 14

Type-in-the-answer (no negative marking) · Algebra · Minima & Maxima · Try it, then check the answer and solution below.

CAT 2025 Slot 2QAAlgebra • Minima & MaximaModerate
If a,b,ca,b,c and dd are integers such that their sum is 46, then the minimum possible value of (a−b)2+(a−c)2+(a−d)2(a-b)^2 + (a-c)^2 + (a-d)^2 is
TITA Answer:
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Answer and solution

Answer: 2

📌 Core Concept
To minimize the expression (a−b)2+(a−c)2+(a−d)2,weneedtomake(a - b)^2 + (a - c)^2 + (a - d)^2 , we need to makeb , c , and dd as close to aa as possible. Given the constraint a+b+c+d=46,wedistributethesumarounda + b + c + d = 46 , we distribute the sum arounda$ to minimize the squared differences.
🔢 Step-by-Step Solution
1
Express the Sum Constraint:
a+b+c+d=46a + b + c + d = 46
2
Express the Objective Function:
S=(a−b)2+(a−c)2+(a−d)2S = (a - b)^2 + (a - c)^2 + (a - d)^2
3
Distribute the Sum Around aa:
Let b=a+x,c=a+y,d=a+zb = a + x , c = a + y , d = a + z.
Then, x+y+z=46−3ax + y + z = 46 - 3a.
4
Minimize SS:
To minimize S,x,y,andS , x , y , andz$ should be as close to 0 as possible.
Since 46−3a46 - 3a must be divisible by 3, aa is approximately 11.511.5. Since aa must be an integer, try a=11a = 11 or a=12a = 12.
5
Case 1: a=11a = 11:
b+c+d=35b + c + d = 35. Distribute as 12,12,1112, 12, 11.
S=(11−12)2+(11−12)2+(11−11)2=1+1+0=2S = (11 - 12)^2 + (11 - 12)^2 + (11 - 11)^2 = 1 + 1 + 0 = 2.
6
Case 2: a=12a = 12:
b+c+d=34b + c + d = 34. Distribute as 11,11,1211, 11, 12.
S=(12−11)2+(12−11)2+(12−12)2=1+1+0=2S = (12 - 11)^2 + (12 - 11)^2 + (12 - 12)^2 = 1 + 1 + 0 = 2.

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