Find the greatest no.of four digits which when divided by 10,11,15 and 22 leaves 3,4,8,15 as remainders, respectively.

As it is given in the question that when we divide the no by 10 it gives remainder 3, So that means the last digit of the no is 3. And we need to find the largest no. LCM of 10,11,15,22=5×2×11×3 =330 Greatest 4 digit number 9999 9999÷330=30 with remainder 99 Greatest 4 digit number divisible by 330 is (9999–99)=9900 Now, just subtract the remainders with the divisor. 10–3=7, 11–4=7, 15–8=7, 22–15=7 Our required number =9900–7 =9893 ans