Q. The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:a)123b)127c)235d)305Correct answer is option 'B'. Can you explain this answer?

Question

Answer

Number 1 = N1 = 1657

Remainder = 6

Number 2 = N2 = 2037

Remainder = 5

Thus,

N1 = 1657-6=1651

N2 = 2037-5=2032

Finding the HCF of both the numbers -

Highest Common Factor (HCF) of two or more numbers is the greatest number that divides each of them exactly.

HCF of 1657 = 13,127

HCF of 2037 = 2,2,2,2,127

The common factor between the two numbers -127

Therefore, the greatest number is 127.

Answer

Answer

Answer:

127

Step-by-step explanation:

Number 1 = N1 = 1657

Remainder = 6

Number 2 = N2 = 2037

Remainder = 5

Thus,

N1 = 1657-6=1651

N2 = 2037-5=2032

Finding the HCF of both the numbers -

Highest Common Factor (HCF) of two or more numbers is the greatest number that divides each of them exactly.

HCF of 1657 = 13,127

HCF of 2037 = 2,2,2,2,127

The common factor between the two numbers -127

Therefore, the greatest number is 127

Answer

Finding the greatest number that leaves a remainder when dividing two numbers requires the use of the Chinese Remainder Theorem. The steps to solve the problem are:

Step 1: Write the two equations in the form of x ≡ a (mod m) and x ≡ b (mod n), where x is the unknown number, a and b are the remainders, and m and n are the divisors.

1657 ≡ 6 (mod m) and 2037 ≡ 5 (mod n)

Step 2: Find the values of m and n that make the two equations solvable simultaneously. The divisors must be coprime (i.e., they have no common factors except 1).

Step 3: Use the Extended Euclidean Algorithm to find the values of r and s that satisfy the equation mr + ns = 1, where m and n are the coprime divisors.

1657 ≡ 6 (mod m) and 2037 ≡ 5 (mod n)

Let m and n be the divisors.

mr + ns = 1

1657 ≡ 6 (mod m) => 1651 ≡ 0 (mod m)

2037 ≡ 5 (mod n) => 2032 ≡ 0 (mod n)

Since m and n are coprime, we can use the Extended Euclidean Algorithm to find r and s.

1651r + 2032s = 1

Step 4: Substitute the values of r and s into the equation x = a*ns*r + b*mr*s and simplify.

x = 6*2032*r + 5*1651*s

x = 13042r + 16935s

Step 5: Find the smallest positive integer solution for x that satisfies the two equations.

Since x is congruent to 6 mod m and 5 mod n, we can use the Chinese Remainder Theorem to find x.

x ≡ 6 (mod m)

x ≡ 5 (mod n)

x = 6 + km

x = 5 + ln

13042r + 16935s = 6 + km

13042r + 16935s = 5 + ln

We can start with k = 0 and l = 1 and find the smallest positive integer solutions for m and n that satisfy the equations.

13042r + 16935s = 6

13042r + 16935s = 5 + n

Subtracting the two equations, we get:

16935s - 16935 = n - 6

16929s = n - 1

Since n is a divisor of 2037, we can try n = 2037 and see if it satisfies the equation.

16929s = 2036

s = 2036/16929 ≈ 0.12

Substituting s into the equation 13042r + 16935s = 6, we get:

13042r + 16935(2036/16929) = 6

13042r ≈ -2016

r ≈ -0.154

Since r and s must be integers, we need to find a multiple of 16935 that is close to 2037. We can try 2037 + 16935 =

Answer

1657-6 and 2837-5
then 1651 and 2032
HCF of 1651and 2032 =127
so answer is b

Answer

Required number = H.C.F. of (1657 - 6) and (2037 - 5)

= H.C.F. of 1651 and 2032 = 127.

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