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1. Find the second smallest positive integer that gives a remainder of 2 when divided by 3 and gives a remainder of 3 when divided by 7.

 

2. When a positive integer is expressed in base 7, it is AB_7, and when it is expressed in base 5, it is BA_5. What is the positive integer in decimal?

 

3. What is the remainder when 1^3+2^3+3^3+...+100^3 is divided by 6?

 

4. There are finitely many primes p for which the congruence \(8x\equiv 1\pmod{p}\) has no solutions x. Determine the sum of all such p.

 Sep 3, 2016

Best Answer 

 #2
avatar+129840 
+5

1. Find the second smallest positive integer that gives a remainder of 2 when divided by 3 and gives a remainder of 3 when divided by 7.

 

We have that

 

N = 3a + 2

N = 7b + 3

 

Subtracting these equations we have that

 

3a - 7b - 1  = 0

 

3a - 7b =  1

 

And the second smallest positive integer results when a = 12 and b = 5

 

So.....the second smallest integer is 38.....just as Max said !!!!

 

 

cool cool cool

 Sep 3, 2016
 #1
avatar+9665 
+5

1) We first have a list of what gives a remainder 2 when divided by 3 and we will have a list of what gives a remainder of 3 when divided by 7.

Gives a remainder 2 when divided by 3:

2,5,8,11,14,17,20,.......

Gives a remainder 3 when divided by 7:
3,10,17,......

Then we add 17 to the LCM of 3 and 7:

17 + 21 = 38

Answer : 38

 Sep 3, 2016
 #2
avatar+129840 
+5
Best Answer

1. Find the second smallest positive integer that gives a remainder of 2 when divided by 3 and gives a remainder of 3 when divided by 7.

 

We have that

 

N = 3a + 2

N = 7b + 3

 

Subtracting these equations we have that

 

3a - 7b - 1  = 0

 

3a - 7b =  1

 

And the second smallest positive integer results when a = 12 and b = 5

 

So.....the second smallest integer is 38.....just as Max said !!!!

 

 

cool cool cool

CPhill Sep 3, 2016
 #3
avatar+9665 
+5

1^3+2^3+3^3+...+100^3 mod 6

= (1+2+3+4+5+6+7+8+........100)^2 mod 6

= 5050^2 mod 6

= 5050 mod 6

= 4.

 Sep 3, 2016
 #4
avatar+9665 
0

This is question 3.

MaxWong  Sep 3, 2016

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