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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.

 Jul 7, 2019
 #1
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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
."

 

In base 10 the last two digits of a make:         a1*9 + 5

In base 10 the last two digits of b make:        5*6 + 3 = 33

In base 10 the last two digits of a-b make:    d1*3 + d0

 

So:  a1*9 + 5 - 33 = d1*3 + d0  or a1*9 - 28 = d1*3 + d0  where 0<= a1 <= 8 and 0<= d1,d0 <= 2

 

Hence possible values for a1*9 - 28 are:  -28    -19    -10    -1    8    17    26    35    44 (just letting a1 run from 0 to 8)

 

The maximum value of d1*3 + d0 is 8, the minimum value is 0, because 0<= d1,d0 <= 2, and a-b can’t be negative, so the only option is d1*3 + d0 = 8

which means that d1 = 2 and d0 = 2. 

 

i.e. the last two digits of a - b in base 3 are 22

.

 Jul 8, 2019
edited by Alan  Jul 8, 2019
edited by Alan  Jul 8, 2019

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