Bunuel wrote:
What is the digit on the units place in the expanded value of \(97^{275} – 32^{44}\)?
A. 1
B. 3
C. 5
D. 7
E. 9
This is the same as finding the units digit of 7ˆ275 - units digit of 2ˆ44;
Number 7 has a cycle of 4 {7,9,3,1}, and Number 2, also, has a cycle of 4 {2,4,8,6};
The units digit of 7ˆ275 will be the cycle number of the set equal to the remainder of 275/4, which is remainder 3, so its cycle num: 3;
The units digit of 2ˆ44 will be the cycle number of the set equal to the remainder of 44/4, which is remainder 0, so its the last cycle num: 6;
So, the units digit of 7ˆ275 - units digit of 2ˆ44 is equal to: 3 - 6 = -3, when we get a negative value, we subtract from 10, thus 10-3=7.
Answer (D).
PS: the cyclicality of a numbers units digit is given by that units digit raised to a power, for instance
[1] has a cycle of 1, because 1ˆ1, 1ˆ2, 1ˆ3… will always return 1;
[2] has a cycle of 4, because after 2 to an exponent divisible by 4 will always repeat the same units digit:
2ˆ1=2 has a units of 2,
2ˆ2=4 has a units of 4,
2ˆ3=8 has a units of 8,
2ˆ4=16 has a units of 6,
2ˆ5=32 has a units of 2 (here it begins to repeat the cycle).
2ˆ6=64 has a units of 4 etc…
[3] has a cycle of 4 {3,9,7,1}
[4] has a cycle of 2 {4,6}
[5] has a cycle of 1 {5}
[6] has a cycle of 1 {6}
[7] has a cycle of 4 {7,9,3,1}
[8] has a cycle of 4 {8,4,2,6}
[9] has a cycle of 2 {9,1}
the end.