Latest TCS Aptitude Question SOLUTION: What is the minimum value of abs(187m – 396n – 526) as m, n take all integer values? Here abs is the absolute value function (that is, if x > 0, then abs(x) = x and if x <

• Placement Papers
• /
• TCS
• /
• Numerical Ability
• /
• Algebra

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40153639) TCS Numerical Ability question Keep an EYE

What is the minimum value of abs(187m – 396n – 526) as m, n take all integer values? Here abs is the absolute value function (that is, if x > 0, then abs(x) = x and if x < 0, then abs(x) = – x).
a. 0
b. 9
c. 2
d. 1

Asked In TCS Garima Srivastava (8 years ago)

Unsolved Read Solution (3)

Is this Puzzle helpful?   (26)   (11) Submit Your Solution Algebra

Read Solution (3) :

aniket gaurav  (8 years ago)

(187m-396n-526) Now the trick is to manipulate the variables
if we take 11 common only from 187m and 396n we can write the expression as
11(17m-36n)-526
Now 17m-36n can be replaced by x as 11(17m-36n)=11x
So 11x-526
Search for a factor of 11 closest to 526 which is 528
Answer is 528-526=2

Is this solution Helpfull? Yes (27) | No   

suresh  (8 years ago)

If |(187m−396n)| is 526 then the given expression attains minimum.
Now observe carefully, both 187, 386 are multiples of 11. So |11(17m−36n)| may not equal to exactly 526 but some value near to 526. Nearest multiple of 11 is 528.
Now |11(17m−36n)|=528
⇒(17m−36n)=48
⇒m=48+36n17
⇒m=2+2n+14+2n17
So for n = 10, we get m = 24.
So |11(17m−36n)| = 528 So minimum value of the given expression is 2.

Is this solution Helpfull? Yes (2) | No   

venkataharnadh  (8 years ago)

187m-396n-526=187(m-2n-3)-22n+35
consider
m-2n-3=1
187(1)-22n+35=222-22n
the value =2 when n=10
m-2n-3=1
n=10
m=24
m,n are integers
so the ans=2

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.