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avatar+140 

a) Show that the sum of 11 consecutive integers is always divisible by 11.

 

b) Show that the sum of 12 consecutive integers is never divisible by 12.

 Apr 10, 2019
 #1
avatar+4322 
+2

a): Label the integers: \(x-5, x-4, x-3, x-2, x-1, x, x+1, x+2, x+3, x+4, x+5=11x\) . And, this number is a multiple of \(11\) , so it is definitely divisible by \(11.\)

 

Try the second one on your own!

 Apr 10, 2019
 #2
avatar+103789 
+1

Thanks, tertre......here's another approach

 

Let the first integer  = x

Let the 11th integer = x + 10

 

Sum  =  [ first integer + 11th integer] * number of terms   / 2     =

 

[ x + x + 10 ] *  11  /  2  =

 

[2x + 10 ] * 11 / 2   =

 

2 [ x + 5 ] * 11 /  2    =

 

[ x + 5 ] * 11       which is divisible by 11

 

 

cool cool cool

 Apr 11, 2019
 #3
avatar+103789 
+1

b)  Similar to the first

 

[ x + x + 11] * 12  / 2  =

 

[2x + 11] * 12 / 2   =

 

[ 2x + 11] * 6  =

 

12x + 66  =

 

[12x + 12(5) ] + 6     

 

12 [ x + 5 ] + 6     which, when divided by 12,  will always leave a remainder of 6

 

 

cool cool  cool

 Apr 11, 2019

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