Let $x$ be the least multiple of $29$ which when divided by $20,21,22,24$ and $28$ leaves the remainders $13,14,15,17$ and $21$ respectively. What is the sum of digits of $x$?

Let's observe the difference between divisors and their respective remainders:$20-13=7$$21-14=7$$22-15=7$$24-17=7$$28-21=7$ Since the difference is constant ($k=7$), the number is of the form:
$$x=\text{LCM}(20,21,22,24,28)\times n-7$$
Let's find the LCM of $20,21,22,24,28$:$20=2^2\times 5$$21=3\times 7$$22=2\times 11$$24=2^3\times 3$$28=2^2\times 7$ $
$\text{LCM}=2^3\times 3\times 5\times 7\times 11=8\times 3\times 5\times 7\times 11=9240$$So,$ x = 9240n - 7$.Wearegiventhat$ x$isamultipleof$ 29$.$$9240n-7\equiv 0(\mathrm{mod}29)$$Le{t}^{\prime }sfind$ 9240 \pmod{29}$:$$9240=29\times 318+18$$So,$ 18n - 7 \equiv 0 \pmod{29} \implies 18n \equiv 7 \pmod{29}$.Testingvaluesfor$ n$:If$ n = 2$:$ 18(2) - 7 = 29$,whichisdivisibleby$ 29$.Thus,$ n = 2$.Nowfind$ x$:$$x=9240\times 2-7=18480-7=18473$$Sumofdigitsof$ x$:$$1+8+4+7+3=23$ $