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In this problem, a and b are positive integers.

When a is written in base 9, its last digit is 5.

When b is written in base 6, its last two digits are 53.

When a - b is written in base 3, what are its last two digits? Assume a - b is positive.

 

 

Note: I have solved the question and included a solution if you are curious.

 Feb 20, 2020
edited by chenxander  Feb 20, 2020

Best Answer 

 #1
avatar+41 
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Oh wait, I got it.

Turn them into modular arithmetic expressions such as \(a\equiv 5\bmod{9}\)  and  \(b\equiv 33\bmod{36}\).

 

Then, convert them into mod 3 and mod 9 expressions in order to get the units and tens digits of the base 3 equivalents of \(a\) and \(b\).

 

\(a\equiv 2\bmod{3}\) --> The first digit of a in base 3 is 2.

\(a\equiv 5\bmod{9}\) --> The value of the two last digits of a in base 3 is 5. Therefore, the second digit of a in base 3 is 1.

 

\(b\equiv 0\bmod{3}\) --> The first digit of b in base 3 is 0.

\(b\equiv 6\bmod{9}\) --> The value of the two last digits of b in base 3 is 6. Therefore, the second digit of b in base 3 is 2.

 

So, we end up with 12 as the last two digits of a and 20 as the last two digits of b (both in base 3).

To subtract, we do 2 - 0 = 2 for the units digit. Since the second digit of a is less than the second digit of b, we carry over 3 from the third digit, which we must assume has a value. Then we do 4 - 2 = 2 for the second digit.

 

The last two digits of a - b in base 3 is 22.

 Feb 20, 2020
 #1
avatar+41 
+2
Best Answer

Oh wait, I got it.

Turn them into modular arithmetic expressions such as \(a\equiv 5\bmod{9}\)  and  \(b\equiv 33\bmod{36}\).

 

Then, convert them into mod 3 and mod 9 expressions in order to get the units and tens digits of the base 3 equivalents of \(a\) and \(b\).

 

\(a\equiv 2\bmod{3}\) --> The first digit of a in base 3 is 2.

\(a\equiv 5\bmod{9}\) --> The value of the two last digits of a in base 3 is 5. Therefore, the second digit of a in base 3 is 1.

 

\(b\equiv 0\bmod{3}\) --> The first digit of b in base 3 is 0.

\(b\equiv 6\bmod{9}\) --> The value of the two last digits of b in base 3 is 6. Therefore, the second digit of b in base 3 is 2.

 

So, we end up with 12 as the last two digits of a and 20 as the last two digits of b (both in base 3).

To subtract, we do 2 - 0 = 2 for the units digit. Since the second digit of a is less than the second digit of b, we carry over 3 from the third digit, which we must assume has a value. Then we do 4 - 2 = 2 for the second digit.

 

The last two digits of a - b in base 3 is 22.

chenxander Feb 20, 2020

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