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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+4609 
+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+128090 
+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+128090 
+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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