Question

Find the least number which when divided by 9,10 and 11 give remainders of 2,3 and 5 respectively.

• A. 1037
• B. 897
• C. 1541
• D. none of these

Answer 1

The correct option is D

none of these

Use Chinese remainder theorem
Let the number be N. N is of the form 9A+2 = 10B+3=11C+5
We will first find the number which is of the form 9A+2=10B+3.-----(1)
For that, reduce (1) to 9A=10B+1.
Find the first integer value of B which satisfies the equation such that A is also an integer.
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 next number which will give a remainder 2 when divided by 9 and a remainder 3 when divided by 10 will be 83+ (9*10) =90
The numbers will fall in an AP with the first digit as 83 and common difference 90 (9*10). The general form of the AP is 83+90k when k=0,1,2...
Now to include the $3^{rd}$ condition( i.e divisor 11 and remainder 5) we can write
83+90k=11C+5 -------(2)
reduce it to
90k+ 78 = 11C
to make it easier cast out 11's from LHS
i.e 2K + 1= 11D
To find the integer solutions
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 this 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.