2007-11-06 13:25:20 · 2 answers · asked by Quagmire77 1 in Science & Mathematics ➔ Mathematics
7
7^2 = 49
7^3 = 343
7^4 = 2401
7^5 = 16807
7 raised to a multiple of 4 ends with 1
7 raised to a multiple of 4 plus 1 ends with 7
7 raised to a multiple of 4 plus 2 ends with 9
7 raised to a multiple of 4 plus 3 ends with 3
5^9 =1953125
1953125/4 = 488281.25
So 5^9 is a multiple of 4 plus 1
So 7^5^9 ends with 7
2007-11-06 14:33:41 · answer #1 · answered by jsardi56 7 · 1⤊ 0⤋
7 follows cyclicity of 4 i.e
7^1 ends with 7
7^2 ends with unit digit 9
7^3 ends with unit digit 3
7^4 ends with unit digit 1
7^5 ends with unit digit 7 again
so there is a difference of 4 in power . every unit digit follow such type of cyclicity. for eg. cyclicity of 4 is 2.
so 5^9=(1+ 4)^9 which by the formula of binomial expansion is 1 + 4(sum real number).
hence unit digit is 7.
2015-05-05 20:04:14 · answer #2 · answered by Aditya 4 · 1⤊ 0⤋