Find three consecutive odd integers such that the largest decreased by 3 times the second is 41 less than...?

Question

the smallest.

(con't)...less than the smallest

Answer 1

Since we have 3 consecutive odd integers,
Let x = the first odd integer. Then (x + 2) and (x + 4) would be the next two odd integers.

"The largest decreased by 3 times the second is 41 less than..."

Less than what? Did you forgot to mention any more info?

Answer 2

enable the 1st integer be x, the 2d be x + 2, and the third be x + 4. enable's symbolize the given suggestions with a marvelous linear equation. So we are going to have: (x + 4) - 3(x + 2) = x - forty seven x + 4 - 3x - 6 = x - forty seven x - x - 3x = -forty seven + 6 - 4 -3x = -forty 5 (Divide the two factors by -3) x = 15 for this reason, the consecutive peculiar integers are 15, 17 and 19. desire this facilitates!

Answer 3

-> The general form of an odd number is 2n-1 or 2n+1
(hope you know that)

So the three consecutive integers could be:
2n+1,2n+3 and 2n+5

Now,the greatest less three times the second is:
(2n+5) - 3*(2n+3)
which is equal to 41 less than the smallest:
(2n+1) - 41

-> (2n+5) - 3*(2n+3) = (2n+1) - 41
-> 2n + 5 - 6n - 9 = 2n + 1 - 41
-> -4n - 4 = 2n - 40
-> 6n = 36
-> n = 6

Now replacing the value of 'n' in 2n+1,2n+3,2n+5....
u get ur answer - 13,15,17

Answer 4

They are 13, 15, and 17.

x, x + 2, x + 4

x + 4 - [3(x + 2)] = x - 41
x + 4 - (3x + 6) = x - 41
-2x - 2 = x - 41
-3x = -39
x = 13
x + 2 = 15
x + 4 = 17

Answer 5

Let the numbers be 2n-1, 2n+1 and 2n+3
Given (2n+3) - 3(2n+1) = (2n-1) - 41
2n - 6n - 2n = -42
-6n = -42
6n = 42, n = 7
So the numbers are 13, 15 and 17

Answer 6

let 3 cons odd nos be x-2,x,x+2
then
((x+2)-3x)+41=x-2
x+2-3x+41+2=x
45-2x=x
45=3x
x=15
so the nos. are
13,15,17