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.

Guest Jul 18, 2018

edited by
Guest
Jul 18, 2018

#1**+1 **

I think that no matter what two numbers you choose, the last 2 digits of "a - b" will be 22 in base 3. Here is an example:

a = 104 in base 10 =125 in base 9, and:

b = 69 in base 10 =153 in base 6, and:

104 - 69 = 35 in base 10 =10**22 in base 3.**

Guest Jul 18, 2018

#2**+1 **

I**n 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.**

**1.**

\(\text{When $a$ is written in base $9$, its last digit is $5$ .}\)

\(\boxed{a \equiv 5 \pmod 9 \qquad \text{ or } \qquad a = 5+9n ,\quad n \in N }\)

**2.**

\(\text{When $b$ is written in base $6$, its last two digits are $53$ .}\)

\(\boxed{b \equiv 33 \pmod {6^2} \quad | \quad 33_{10} =53_6\\ \qquad \text{ or }\\ b = 33+ 36m ,\quad m \in N }\)

**3.**

\(\text{When $a-b$ is written in base $3$, what are its last two digits?}\)

\(\begin{array}{|rcll|} \hline a &=& 5+ 9n \\ b &=& 33+36m \\ \hline a-b &=& 5+9n -(33+36m) \\ &=& -28 + 9n - 36m \\\\ && -28 + 9n - 36m \pmod{3^2} \quad & | \quad \text{last two digits in base }3 \\ &=& -28 + 36 +9n - 36 m \pmod{9} \\ &=& 8 +9n - 36 m \pmod{9} \\ &=& 8 +0 - 0 \\ &=& 8_{10} \\ &=& 22_{3} \\ \hline \end{array} \)

The last two digits in base 3 are **22**.

heureka Jul 19, 2018