find the sum of all 3-digit numbers which are not multiples of 6

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bumblebee123

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can anyone help me?

question: find the sum of all 3-digit numbers which are not multiples of 6

I've started by attempting to find the sum of all 3-digit numbers which are multiples of 6

Sn = n/2 [ 204 + ( n-1 )6 ]

Sn = 198 + 6n

I have no idea what to do next!
 
Yes and what actually is the highest 3-digit number which is a mutiple of 6? Is it 999, 998,....??? You can then use this to find n, and then go back to your first post to find the sum. (Note that in your first post you lost the n/2 on your second line.)
 
Harry_the_cat said:
Yes and what actually is the highest 3-digit number which is a mutiple of 6? Is it 999, 998,....??? You can then use this to find n, and then go back to your first post to find the sum. (Note that in your first post you lost the n/2 on your second line.)
Click to expand...

For some reason,I thought the amount of 3-digit multiples of 6 was infinite :rolleyes:

but you're right- obviously, it'll turn into a four digit number at some point. the highest 3-digit multiple of 6 = 996

996 = 96 + 6n
6n = 900
n = 150

S150 = 75 ( 198 + 444 )

S150 = 48150
 
S150 = n/2 ( a + last term )
S150 = 75 x ( 102 + 996 )
S150 = 82350

what have I done wrong?
 
996 = 96 + 6n
6n = 900
n = 150

n/2 ( 198 + 6n )

S150 = 75 x ( 198 + ( 6 x 150 ) )
S150 = 75 x ( 198 +900 )
S150 = 75 x 1098
S150 = 82350
 
so the sum for 3-digit numbers which are not multiples of 6 = sum of all three-digit numbers - 82350
 
the sum for 3-digit numbers which are not multiples of 6 = sum of all three-digit numbers - 82350
 
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