A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. What is the least such number?
Answer: A
Let the number be N. When N is divided by 4, let the quotient be x and remainder be 1. So, \(N = 4x + 1\). Now, this quotient x is divided by 5, leaving a remainder of 4. So, \(x = 5y + 4\). To find the least number, we take the least possible value for the final quotient y, which is 0. No, we must take y=1 to find the second quotient. Let's start from the end. Let the second quotient be k. So, \(x = 5k + 4\). For the least number, we take the smallest possible value for k, which is 1. So \(x = 5(1)+4=9\). Now substitute x back: \(N = 4(9) + 1 = 36 + 1 = 37\). Wait, let's try with k=0. Then \(x = 5(0)+4=4\). Then \(N = 4(4)+1=17\). Let's check 17. \(17 \div 4 = 4\) rem 1. Correct. The quotient is 4. Now divide 4 by 5. \(4 \div 5 = 0\) rem 4. Correct. So 17 is the least number.