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  1. The largest number which divides 30, 78 and 102 to leave the same remainder in each case is

Options:
A .  8
B .  16
C .  20
D .  24
Answer: Option D
To solve this problem, we need to find the greatest common divisor (GCD) of the three given numbers. The GCD is the largest number that divides all the given numbers without leaving any remainder. However, in this case, we are also given that the number should leave the same remainder when it divides each number.
Here are the steps to solve the problem:
Step 1: Find the difference between any two of the given numbers.Difference between 30 and 78 = 78 - 30 = 48Difference between 78 and 102 = 102 - 78 = 24Difference between 102 and 30 = 102 - 30 = 72
Step 2: Find the GCD of the differences.GCD(48, 24, 72) = 24
Step 3: The GCD of the differences is the required answer.Therefore, the largest number which divides 30, 78 and 102 to leave the same remainder in each case is 24.
Explanation:
  • GCD: The greatest common divisor (GCD) of two or more integers is the largest positive integer that divides each of the integers without leaving a remainder.
  • Remainder: The remainder is the amount left over after division. For example, when we divide 7 by 3, the quotient is 2 with a remainder of 1.
  • Difference: The difference between two numbers is the result of subtracting one number from the other.
  • Euclidean Algorithm: The Euclidean algorithm is a method for finding the GCD of two numbers. The algorithm states that the GCD of two numbers is equal to the GCD of the smaller number and the remainder of the larger number divided by the smaller number.
In this problem, we used the fact that if a number leaves the same remainder when it divides three given numbers, then it must also leave the same remainder when it divides the differences between those numbers. By finding the GCD of the differences, we were able to find the GCD of the original numbers. The final answer is 24, which is the largest number that satisfies the given conditions.

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5 Comments

HCF OF [(78-30)(102-78)(102-30)]
HCF OF [(48)(24)(72)]
48=2^4.3
24=2^3.3
72=2^3.3^2
HCF=2^3.3=24
(30-r)-(78-r) and (102-r)-(30-r)
-48 and 72
-48 and 72 H.C.F 24
102-78,102-30,78-30 =24,72,48
hcf of 24,72,48 is 24
24
=> let the number be n which divides 30, 78 and 102 leaving a reminder r

=> the required number then becomes H.C.F of 30-r, 78-r and 102-r

=> it could also be the H.C.F of 
(78-r)-(30-r) and (102-r)-(78-r)

i.e. 48 and 24

=> H.C.F of 48 and 24 = 24

the required number n is 24.

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