plz tell quickly
2007-05-25 02:31:52 · 10 answers · asked by galin 2 in Science & Mathematics ➔ Mathematics
23214332421309129032130821382130921832103213802913
this is the number actually sme numbers r nt shoen
2007-05-25 02:40:27 · update #1
232143324213091290321308213821
2007-05-25 02:43:29 · update #2
23214332421 3091290321 308213821
2007-05-25 02:45:30 · update #3
It is not even number it can be divisible and has factors
2007-05-25 02:35:17 · answer #1 · answered by Mr.Karachi 5 · 0⤊ 0⤋
Not if the number ends in 2.
2007-05-25 09:36:24 · answer #2 · answered by kyle v 2 · 0⤊ 2⤋
It's a composite number, with the prime factors:
(11)(73)(289095048833239464908229407)
2007-05-25 09:54:05 · answer #3 · answered by Scythian1950 7 · 1⤊ 0⤋
THE OTHERS NEED TO RE-LOOK AT THE NUMBER AND SEE THAT IT ENDS WITH A "1," NOT A "2." THIS NUMBER IS DEFINITELY A PRIME NUMBER.
2007-05-25 09:53:45 · answer #4 · answered by big d 3 · 0⤊ 0⤋
If your entire number is not visible, please put a space between every 3 digits so we can see all of it and advise you correctly.
e.g.
232 143 324 213 091 290 321 308 213 821 ...
2007-05-25 09:43:31 · answer #5 · answered by MamaMia © 7 · 0⤊ 0⤋
No - it's even. The only even prime number is 2.
To be prime the number can only be divisible by 1 and itse;f. Any even number is divisible by 2 as well.
2007-05-25 09:36:16 · answer #6 · answered by Isaac H 3 · 0⤊ 2⤋
It is divisible by 17.
2007-05-25 09:48:29 · answer #7 · answered by wadi 2 · 0⤊ 0⤋
no
for sure divisible by 11
2007-05-25 09:58:30 · answer #8 · answered by yerffej89 3 · 0⤊ 0⤋
no
2007-05-25 09:50:54 · answer #9 · answered by kimballclay 2 · 0⤊ 0⤋
no
2007-05-25 09:35:57 · answer #10 · answered by nerd.:) 4 · 0⤊ 2⤋