Find two consecutive odd integers whose sum is 116?
2006-09-12 17:16:07 · 11 answers · asked by Hiphopaction 2 in Science & Mathematics ➔ Mathematics
Simplest way to do this:
116/2 = 58
then:
58 +/- 1 = 57, 59; 57+59 = 116
2006-09-12 17:18:43 · answer #1 · answered by Sean06 2 · 2⤊ 1⤋
57 n 59
2006-09-13 00:41:26 · answer #2 · answered by confused 2 · 1⤊ 0⤋
let the intergers be
x and x+2
hence the sum is,
x + (x + 2) = 116
2x + 2 = 116
2(x + 1) = 116
x + 1 = 116/2
x = 58 - 1
x = 57
Hence, the numbers are 57 and 59.
2006-09-13 00:36:07 · answer #3 · answered by ideaquest 7 · 1⤊ 0⤋
Ok, i believe the easiest way to solve this problem is:
116 : 2 = 58
(58 - 1) + (58 + 1) = 116
57 + 59 = 116
Is this the answer you need?
2006-09-13 00:28:11 · answer #4 · answered by becko 1 · 2⤊ 0⤋
+57 and +59
2006-09-13 00:25:28 · answer #5 · answered by marvs:) 1 · 1⤊ 0⤋
57&59
2006-09-13 00:20:40 · answer #6 · answered by Scott S 4 · 1⤊ 0⤋
x + (x + 2) = 116
x + x + 2 = 116
2x + 2 = 116
2x = 114
x = 57
ANS : 57 and 59
2006-09-13 01:39:11 · answer #7 · answered by Sherman81 6 · 1⤊ 0⤋
57&59, BANK ON IT
2006-09-13 00:23:51 · answer #8 · answered by kidfrom68 1 · 1⤊ 0⤋
She said odd integers you dumb as s
2006-09-13 00:24:50 · answer #9 · answered by big10inmidd 2 · 0⤊ 1⤋
58 unless its in negatives
2006-09-13 00:21:43 · answer #10 · answered by kacak15348 1 · 0⤊ 1⤋