2006-10-04 08:35:54 · 8 answers · asked by LE 1 in Science & Mathematics ➔ Mathematics
So if x is the smallest of the 3, you're saying..
x+2 (the second number) + x+4 (the third number) = x+23
...
x+2+x+4 = x+23
2x+6=x+23
x= 23-6=17
To check, 19+21 = 40, which is 23 bigger than 17
2006-10-04 08:39:59 · answer #1 · answered by Sean M 1 · 0⤊ 0⤋
Take 3 consecutive integers as x , x+a million , x+2. (x+a million) + (x + 2) = 24 + (a million/2)x 2x+3 = 24 + (a million/2)x 2x+3 = (40 8 + x)/2 multiply each and each aspect by skill of two. 4x + 6 = 40 8 + x 3x = 40 2 x = 14 the three consecutive integers are 14,15,16 Have an wonderful day. :)
2016-11-26 02:42:55 · answer #2 · answered by Anonymous · 0⤊ 0⤋
17 19 21
In general terms, if you are to find three consecutive odd integers such that the sum of the second and third are a value z greater than the first, then the third integer must be (z-2)
2006-10-04 12:40:04 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Let 2n+1 be the first odd integer. Then the next ones are 2n+3 and 2n+5.
The sum of the 2nd and 3rd is 4n+8, so if that is 23 greater than the first, you have
4n+8=2n+1+23=2n+24, so 2n=16, meaning n=8, so your integers are 17, 19 and 21.
2006-10-04 08:39:43 · answer #4 · answered by James L 5 · 1⤊ 0⤋
x=first integer
x+2 = seond integer
x+4 = third integer
x+2+x+4=23+x
2x+6=x+23
x=17
17, 19, 21
2006-10-04 08:47:20 · answer #5 · answered by T 5 · 0⤊ 0⤋
17, 19, 21
2006-10-04 08:41:31 · answer #6 · answered by johannsinuhe 2 · 0⤊ 0⤋
17,19,21
2006-10-04 08:43:19 · answer #7 · answered by rameezaali 2 · 0⤊ 0⤋
17,19,21
2006-10-04 08:39:56 · answer #8 · answered by Greg G 5 · 0⤊ 0⤋