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complete \(2011 \cdot 2012 \cdot 2013 \cdot 2014 \) modulo 5.

 

and

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.

lololololololololll  May 20, 2018
 #1
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(2011*2012*2013*2014) mod 5 = 4

Guest May 20, 2018
 #2
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What about 2?

lololololololololll  May 20, 2018
 #3
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You can have an infinite number of solutions:

We can choose "a" to be 140 in base ten = 165 in base 9

We can choose "b" to be 105 in base ten = 253 in base 6

 

Now, we can convert 140 and 105 from base 10 to base 3:

140 in base 10 =1 2012 in base 3, and:

105 in base 10 =1 0220 in base 3. So:

1 2012 - 1 0220 =1022 in base 3.

Guest May 20, 2018

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