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1)  Determine the smallest non-negative integer  that satisfies the congruences: 

a = 2 (mod 3)

a = 4 (mod 5)
a = 6 (mod 7)

a = 8 (mod 9)

 

2)  What is the average of all positive integers that have four digits when written in base 3, but two digits when written in base 6? Write your answer in base 10.

 

Q3 deleted.

Just ask one question per post. Thanks.

 May 26, 2019
edited by Melody  May 27, 2019
 #1
avatar+4325 
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1): We can use the Chinese Remainder Theorem.

 

I can post a solution if you want.

 May 26, 2019
 #2
avatar+104962 
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2)  What is the average of all positive integers that have four digits when written in base 3, but two digits when written in base 6? Write your answer in base 10.

 

The integers in base 3 can range from

 

10003 to  22223   =   [27  to 80] base10

 

The integers in base 6 can range from

 

106  to  556  = [6 to 35] base 10

 

So  the common integers are      27, 28, 29 , 30 , 31, 32, 33 , 34 and 35

 

Their average  =   [ 4(62) + 31]  / 9  =  31

 

 

cool cool cool

 May 26, 2019
edited by CPhill  May 26, 2019
 #3
avatar+4325 
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I might as well post a solution to the first problem...

 

 

a = 2 (mod 3)

a = 4 (mod 5)
a = 6 (mod 7)

a = 8 (mod 9)

 

The above is the system of congruences.

 

The answer is 314, by using the general Chinese Remainder Theorem. https://www.youtube.com/watch?v=xjtx4CaQa_4

 May 26, 2019
 #4
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Please just help with problems 1 and 3. Thank you for helping with problem 2

 May 26, 2019
 #5
avatar+105704 
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Thanks Tertre and Chris.

 May 27, 2019
edited by Melody  May 27, 2019

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