Find the least number which when divided by 9,10 and 11 give remainders of 2,3 and 5 respectively?
Find the least number which when divided by 9,10 and 11 give remainders of 2,3 and 5 respectively?
The correct option is D None of these
Use Chinese remainder theorem
Let the number be N. N is of the form
9A + 2(i.e. N divided by 9 gives a remainder 2) = 10B + 3 (i.e. N divided by 10 gives a remainder 3) = 11C + 5 (i.3. N divided by 11 gives a remainder of 5)
We will first find the number which is of the form
9A + 2 = 10B + 3. ------------------------- (1)
Here for B = 8, A = 9, which are the first integer solutions. Substituting this in equation (1) we get the value 83, the first number which when divided by 9 gives a remainder 2 and which when divided by 10 gives a remainder 3.
The numbers will fall in an AP with the first digit at 83 and common difference 90 (9 * 10). The general form of AP is 83 + 90k when k = 0, 1, 2….
Now to include the 3rd condition (i.e. divisor 11 and remainder 5) we can write
83 + 90k = 11c + 5……(2)
Find the first integer value of K which satisfies the equation such that c is also an integer
Here for k = 5, c = 4, which are the first integer solutions.
(To make equation (2) simpler, cast out the 11s… the equation changes to 2k + 1 = 11D. Find the first integer value of K which satisfies the equation such that D is also an integer.
Here for K = 5, D = 1, which are the first integer solutions substituting the value of K in equation
(2) we get the value 90 (5) + 83 = 533, the first number which when divided by 9 gives a remainder 2, which when divided by 10 gives a remainder 3 and which when divided by 11 gives a remainder 5)
None of these
Use Chinese remainder theorem
Let the number be N. N is of the form
9A + 2(i.e. N divided by 9 gives a remainder 2) = 10B + 3 (i.e. N divided by 10 gives a remainder 3) = 11C + 5 (i.3. N divided by 11 gives a remainder of 5)
We will first find the number which is of the form
9A + 2 = 10B + 3. ------------------------- (1)
Here for B = 8, A = 9, which are the first integer solutions. Substituting this in equation (1) we get the value 83, the first number which when divided by 9 gives a remainder 2 and which when divided by 10 gives a remainder 3.
The numbers will fall in an AP with the first digit at 83 and common difference 90 (9 * 10). The general form of AP is 83 + 90k when k = 0, 1, 2….
Now to include the 3rd condition (i.e. divisor 11 and remainder 5) we can write
83 + 90k = 11c + 5……(2)
Find the first integer value of K which satisfies the equation such that c is also an integer
Here for k = 5, c = 4, which are the first integer solutions.
(To make equation (2) simpler, cast out the 11s… the equation changes to 2k + 1 = 11D. Find the first integer value of K which satisfies the equation such that D is also an integer.
Here for K = 5, D = 1, which are the first integer solutions substituting the value of K in equation
(2) we get the value 90 (5) + 83 = 533, the first number which when divided by 9 gives a remainder 2, which when divided by 10 gives a remainder 3 and which when divided by 11 gives a remainder 5)
