When 93, 119 and 158 are divided by the integer n(n > 1) , the remaind

1.	When 93, 119 and 158 are divided by the integer n(n > 1) , the remainder in each case is d. What is the value of (n+d)?1. 172. 193. 144. 15
Answer» Correct Answer - Option 4 : 15 Given : 93, 119 and 158 when divided by n, give remainder d in each case Concept used : HCF of 3 different numbers a, b and c = HCF of (b -a), (c -b) and (c -a) Calculations : 93, 119 and 158 gives remainder d when divided by n Then, (93 - d), (119 -d) and (158 - d) must be divisible by n Now HCF of (93 - d), (119 - d) and (158 - d) = HCF of (119 - d - (93 - d), (158 -d -(119 -d) and (158 - d - (93 -d) ⇒ HCF of 26, 39 and 65 ⇒ 13 So value of n will be 13 Remainder we get on dividing 93, 119 and 158 by 13 is 2 So d = 2 S0, n + d = 13 + 2 ⇒ 15 ∴ The value of (n + d) is 15

When 93, 119 and 158 are divided by the integer n(n > 1) , the remainder in each case is d. What is the value of (n+d)?1. 172. 193. 144. 15

Answer» Correct Answer - Option 4 : 15

Given :

93, 119 and 158 when divided by n, give remainder d in each case

Concept used :

HCF of 3 different numbers a, b and c = HCF of (b -a), (c -b) and (c -a)

Calculations :

93, 119 and 158 gives remainder d when divided by n

Then, (93 - d), (119 -d) and (158 - d) must be divisible by n

Now

HCF of (93 - d), (119 - d) and (158 - d) = HCF of (119 - d - (93 - d), (158 -d -(119 -d) and (158 - d - (93 -d)

⇒ HCF of 26, 39 and 65

⇒ 13

So value of n will be 13

Remainder we get on dividing 93, 119 and 158 by 13 is 2

So d = 2

S0,

n + d = 13 + 2

⇒ 15

∴ The value of (n + d) is 15

Discussion

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