Question

Find the largest four-digit number which when divided by 4, 7 and 13 leaves a remainder of 3 in each case.

Answer 1

Prime factors of $$4, 7$$ and $$13$$
$$ 4 = 2 \times 2 $$
$$7$$ and $$13$$ are prime numbers
LCM $$( 4, 7 , 13)= 364$$
we know that, the largest $$4$$ digit number is $$9999.$$
Step 1 : divide $$9999$$ by $$364 ,$$ we get
$$\dfrac {9999} {364} = 171$$
Step 2: subtract $$171$$ from $$9999$$
$$9999-171 = 9828$$
Since a remainder of $$3$$ is to be left
$$9282 +3 = 9831$$
therefore $$9831$$ is the number.