Ok lets say you hafta find three consecutive EVEN integers for 147
You would do:
n+n+2+n+4=147
then
3n+6=147
then subtract six
3n=141
then divide 3
n=47
ok so 47 isnt even, would you just go up one and write 48, 50, and 52? Help please!!
2007-12-05 12:45:13 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
If you are told that the sum must *equal* 147, then there is nothing you can do.
If you are told that the sum must be as big as possible, but no bigger than 147, then go down.
If you are told that the answer must be 147 or bigger, then go up.
But basically this sounds like a trick question.
All your math is correct, but the question is contradictory.
2007-12-05 12:53:49 · answer #1 · answered by Puzzling 7 · 1⤊ 0⤋
n+(n+2)+(n+4)=147
3n+6=147
-6 from both sides
3n=141
both sides divided by 3
47
the three consecutive numbers can't be 48,50,52 because that doesn't add up to 147(it equals 150)
This problem is impossible because no two (or three)even numbers can add up to 7.
But I may be wrong, maybe there are three numbers adding to 7.
2007-12-05 12:56:26 · answer #2 · answered by On Earth 3 · 0⤊ 0⤋
including. adverse plus adverse is adverse. (–18) + (–19) + (–20) = –57 Oops! attempt –17, –18 and –19! x + (x + a million) + (x + 2) = 3x + 3 = –fifty 4, 3x = –57, x = –19 (note that because adverse numbers advance in importance to the left, x + a million = –18 and x + 2 = –17)
2016-10-26 13:21:02 · answer #3 · answered by ? 4 · 0⤊ 0⤋
your math is right but... the sum of any even integers can NOT be a negative number.
And no, you can't just go up to 48, 50, 52 becauce 48+50+52 is not equal to 147
Rec
2007-12-05 12:50:26 · answer #4 · answered by Anonymous · 2⤊ 1⤋
It's impossible. 3 even integers will ALWAYS add up to an even number, which 147 isn't. You solved it perfectly, but the question is impossible. The answer IS 47,49,51, but those aren't even, so there's no answer.
2007-12-05 12:48:56 · answer #5 · answered by SaintPretz59 4 · 3⤊ 0⤋
This problem could be done if it was 3 ODD consecutive inters or 3 POSITIVE consecuitive interers, yet it is imposible with 3 EVEN consecutive intigers
2007-12-05 12:55:13 · answer #6 · answered by Jason Frost 3 · 0⤊ 1⤋
Your problem has an error in it, your work was done correctly.
Don't ever just round up to make it work.
If the problem was ODD integers, everthing is perfect!
2007-12-05 12:51:11 · answer #7 · answered by RickSus R 5 · 2⤊ 0⤋
there is no answer, since even+even is always even
2007-12-05 12:59:43 · answer #8 · answered by Zack C 3 · 0⤊ 0⤋