Math Challenge?

1) First notice that 3456/8=432, and so these are the first 4 digits of the quotient. As we know the remainder is 5, we know that 3456N2 is a multiple of 8. We know 3456 is a multiple 8, so N2 is a multiple of 8. This means N=3 or 7

2) Let n be the number we want to find. Then the numbers in the sequence are n-6, n-4, n-2, n. The sum of these is 4n-12.
The average of these is n-3. We know the average = 17, so n=20

1. 3 and 7

2. 20. (Avg of 20, 18, 16, 14 is 17)

1) The value of N is 3 or 7

2) Let n be the largest of 4 consecutive integers

n + (n - 2) + (n - 4) + (n - 6) = 4 * 17
n + (n - 2) + (n - 4) + (n - 6) = 68
4n - 12 = 68
4n = 80
n = 20

1) 3 and 7

2) 20

1) N = 4,9
2) 20

Hi,

2. First

Let x = the smallest consecutive integer
Then x + 2 = next smallest
x + 4 = third smallest
x + 6 = largest
so
(x + x + 2 + x + 4 + x + 6)/4 = 17
then (4x + 12) / 4 = 17 sp x + 3 = 17 then x = 14 then
x + 6 = 20.

1. n = 3 or n = 7 So 4*8 = 32 form 37 then subtracting gives 5 for a remainder

Also 7 so we would 77 then 9*8 = 72 and when we subtract we have 5.

Hope This Helps!!