# Thread: Quotient and remainder.

1. ## Quotient and remainder.

find the quotient and remainder when 11,109,999,999,997 is divided by 1111.

convert (1100101)2 to base 10 convert 32,145 to octal??????

the sum of the digits of the number 8215 is 8+2+1+5=16=7(mod9). 8215=9(912)+7=7(mod 9). Does this hold for any number?

2. Hello, tygracen!

Here are a few of them . . .

1. Find the quotient and remainder when 11,109,999,999,997 is divided by 1111.

$11,\!109,\!999,\!999,\!997 \:=\:11,\!110,\!000,\!000,\!000 -3$

Then we have: . $\frac{11,\!110,\!000,\!000,\!000 - 3}{1111}$

Subtract and add 1111:

. . $\frac{11,\!110,\!000,\!000,\!000 \;{\color{red}- \:1111} - 3 {\color{red}\:+\:1111}}{1111} \;=\;\frac{11,\!110,\!000,\!000,\!000,\!000 - 1111 + 1108}{1111}$

. . $= \;\frac{11,\!110,\!000,\!000,\!000}{1111} - \frac{1111}{1111} + \frac{1108}{1111} \;=\;10,\!000,\!000,\!000 -1 + \frac{1108}{1111}$

. . $= \;9,\!999,\!999,\!999 + \frac{1108}{1111}$

The quotient is 9,999,999,999 . . . the remainder is 1108.

2. Convert $1,\!100,\!101_2$ to base 10.

$1,\!100,\!101_2 \;=\;1\!\cdot\!2^6 + 1\!\cdot\!2^5 + 0\cdot\!2^4 + 0\cdot\!2^3 + 1\!\cdot\!2^2 + 0\cdot\!2 + 1 \;= \;64 + 32 + 4 + 1 \;=\;101$

3. Convert 32,145 to octal.

$32,\!145 \;=\;7\!\cdot\!8^4 + 6\!\cdot\!8^3 + 6\!\cdot\!8^2 + 2\!\cdot\!8 + 1 \;=\;76,\!621_8$