possible solutions...
1238
16586546893354754657
1
1239
or is it none of these
2007-05-06 15:59:01 · 9 answers · asked by Olivia 4 in Science & Mathematics ➔ Mathematics
The definition of a 1239! is:
1239 x 1238 x 1237 x .... x 3 x 2 x 1
Likewise, 1238! = 1238 x 1237 x 1236 x .... x 4 x 3 x 2 x 1
So... when dividing 1239! by 1238! , try seeing which terms cancel out. Your answer is what's left.
2007-05-06 16:05:36 · answer #1 · answered by hanyo_dossta 2 · 2⤊ 0⤋
The answer is 1239. Think about it- it's more of a logical question than anything.
1239! is 1239*1238*1237*1236...*1
What is 1238!?
Well it is 1238*1237*1236...*1
What is the only number that is not included in 1238! that is included in 1239!?
It's 1239 of course. The rest of the numbers all cancel each other out when dividing.
2007-05-06 23:10:09 · answer #2 · answered by Kathryn S 4 · 0⤊ 0⤋
the answer is 1239 because
1239! = 1239 * 1238!
so
1239 * 1238! / 1238!
= 1239
2007-05-06 23:07:11 · answer #3 · answered by rooster1981 4 · 0⤊ 0⤋
1239
2007-05-06 23:06:12 · answer #4 · answered by ignoramus 7 · 0⤊ 0⤋
1239*1238!/1238! = 1239
2007-05-06 23:03:35 · answer #5 · answered by fcas80 7 · 1⤊ 0⤋
1239
2007-05-06 23:03:33 · answer #6 · answered by Maan_palaboy 2 · 0⤊ 1⤋
It is 1239
1239!/1238! =1239*1238......*1/1238*1237*....*1
This obviously cancels out everything except 1239 which is your answer.
2007-05-06 23:11:36 · answer #7 · answered by Gully 2 · 1⤊ 0⤋
definitely 1239
2007-05-06 23:08:56 · answer #8 · answered by jimj21 3 · 0⤊ 0⤋
1239 is the correct answer
ex. 6!/5!
= (6*5*4*3*2*1)/(5*4*3*2*1)
as you can see they will all cancel except for the six hope this helps..
2007-05-06 23:07:30 · answer #9 · answered by Scuba Dave 1 · 0⤊ 0⤋