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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.

Guest Jul 18, 2018
edited by Guest  Jul 18, 2018
#1
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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 =1022 in base 3.

Guest Jul 18, 2018
#2
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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.

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