Write the number 112 in "expanded form."

The expanded form in any x ( x is an imteger) base system is like a polynomial in the form:

An*x^n+An-1x^(n-1)+An-2*x^(n-2)+An-3*x^(n-3)+......A2*x^2+A1*x+A0, where An , An-1, An-2, ....A2,A1 and A0 are the positive integral coefficients such that 0 <= Ar< x and  r is  an integer 1 to n.

You did not cite the base. So, we take first 10 base , the most popular a decimal system and then  one hundred and twelve converted in base 5   and its expansion in the powers of 5 and   the  number 112 in base 5 and its expansion.

Procedure:

(i)

112 in 10 base:

So the 112/10 = quotient 11+remainder 2.

Take the quotient 11 above and divide by 10 to get the quotient and remainder:

11/10= quotient 1+ remainder 1

Take the quotient 1 above and again divide to get quotient and remainder:

1/10 = quotient 0 + remainder 1

Procedure is to go till the quotient is 0.

112= 3rd remainder* 10^ (3-1)+(second remainder*10^(2-1)+first remaider*10^(1-1)'

Therefore, 112= 1*10^(3-1)+1*10^(2-1)+2*10^(1-1) or

112=1*10^2+1*10+2*10^0 or

112=1*10^2+1*10^1+2*10^0 0r

112=1*100+1*10+2.

(ii)

112 converted in 5 base:

112/5=quotient 22 +remainder 2

22/5= quotient 4 + remainder 2

4/5= quotient 0  remainder 4

Therefore 112 (one hundred and  five) in base 5 =  4*5^2+2*5+2.

(iii)

Now 112 in base 5 = 1*5^2+1*5+2