# Quotient and remainder.

• Dec 2nd 2008, 05:47 PM
tygracen
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?
• Dec 2nd 2008, 07:32 PM
Soroban
Hello, tygracen!

Here are a few of them . . .

Quote:

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

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

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

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

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

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

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

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3. Convert 32,145 to octal.

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