Also, the originally problem is finding the ones digit 537^102
and I found out that 7 was a one digit factor
7^0 = 1
7^1 = 7
7^2 = 49
7^3 = 343
7^4 = 2401
Pattern of ones digit is 1,7,9,3,1,7,9,3...
However, I don't understand how you can find this pattern in the values given above?
5 answers
What the patters is supposed to show is..
in 537^102 you only have to look at what happens to the ones digit
e.g.
537^0 = 1 vs 7^0 = 1
537^1 = 537 vs 7^1 = 77
537^2 = 288369 vs 7^2 = 49
537^3 = 157854153 vs 7^3 = 343
537^4 = 8315....1 vs 7^4 = 2401
notice the last digit is either
1 , 7, 9, or 3 , and cycles in that pattern
If the exponent is odd ---> not our problem
if the exponent is even and divides by 4, then it end in a 1
if the exponent is even but does not divide by 4, it ends in a 9
our exponent of 102 is even but not divisible by 4, so
537^102 will end in a 9
in 537^102 you only have to look at what happens to the ones digit
e.g.
537^0 = 1 vs 7^0 = 1
537^1 = 537 vs 7^1 = 77
537^2 = 288369 vs 7^2 = 49
537^3 = 157854153 vs 7^3 = 343
537^4 = 8315....1 vs 7^4 = 2401
notice the last digit is either
1 , 7, 9, or 3 , and cycles in that pattern
If the exponent is odd ---> not our problem
if the exponent is even and divides by 4, then it end in a 1
if the exponent is even but does not divide by 4, it ends in a 9
our exponent of 102 is even but not divisible by 4, so
537^102 will end in a 9
Thank you so much! Your explanation was very helpful!
the 2nd line of the e.g.'s should say
537^1 = 537 vs 7^1 = 7
537^1 = 537 vs 7^1 = 7
7 xxxxx
1 answer