Hello, Probability!
Your question isn't clear . . .
Your example is unfortunate . . .
We are given 19,375.
(a) Do we already know that it is the sum of powers-of-5?
If so, divide it by 5 repeatedly as long as it "comes out even".
. . \(\displaystyle \begin{array}{ccc} 19375 \div 5 &=& 3875 \\ 3875 \div 5 &=& 775 \\ 775 \div 5 &=& 155 \\ 155 \div 5 &=& 31 \end{array}\)
We find that: \(\displaystyle 19,\!375 \:=\:5^4\cdot31\)
Then we find that: \(\displaystyle 31 \:=\:1+5+25 \:=\:1+5+5^2\)
Therefore: .\(\displaystyle 19,\!375 \:=\:5^4(1 +5+5^2) \:=\:5^4 + 5^5 + 5^6\)
(b) Are you converting 19,375 to a base-five number?
. . .If so, there is a procedure for that.
[1] Divide the number by 5; note the remainder.
[2] Divide the quotient by 5; note the remainder.
[3] Repeat step [2] until a zero quotient is attained.
[4] Read up the remainders.
\(\displaystyle \begin{array}{ccccccc}19375 \div 5 &=& 3875 & \text{rem. 0} \\ 3875 \div 5 &=& 775 & \text{rem. 0} \\ 775 \div 5 &=& 155 & \text{rem. 0} \\ 155 \div 5 &=& 31 & \text{rem. 0} \\ 31 \div 5 &=& 6 & \text{rem. 1} \\ 6 \div 5 &=& 1 & \text{rem. 1} \\ 1 \div 5 &=& 0 & \text{rem. 1} \end{array}\)
Therefore: .\(\displaystyle 19,\!375 \:=\:1,\!110,\!000_5\)
