The questions tests the knowledge of cyclisity.
Here we are given 2 number but our main focus should be on the last digits of both the numbers, that is, 7 and 2.
Cyclisity of 7 is 4 (7, 9, 3, 1), so in \(7^{275}\), when we divide 275 by 4 and get 3 as a remainder. So the number \(7^{275}\) will end with the last digit of \(7^{3}\) and that is 3.
On the similar lines, the Cyclisity of 2 is 4 (2, 4, 8, 6), so in \(2^{44}\), when we divide 44 by 4 and get 0 as a remainder. When the remainder = 0, the number ends in \(2^{cyclisity}\) = \(2^{44}\), that is 6.
Now we have (Number ending with 3) - (Number ending with 6).Think of a few examples :
13 - 6 = 7 /
23-16 = 7 /
143-56 = 87 (Units digit is 7).
So, the last digit would be 7. Answer is D.
Relevant Theory -
Quote:
LAST DIGIT OF A POWER
Determining the last digit of \((xyz)^n\):
1. Last digit of \((xyz)^n\) is the same as that of \(z^n\);
2. Determine the cyclicity number \(c\) of \(z\);
3. Find the remainder \(r\) when \(n\) divided by the cyclisity;
4. When \(r>0\), then last digit of \((xyz)^n\) is the same as that of \(z^r\) and when \(r=0\), then last digit of \((xyz)^n\) is the same as that of \(z^c\), where \(c\) is the cyclisity number.
• Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base.
• Integers ending with 2, 3, 7 and 8 have a cyclicity of 4.
• Integers ending with 4 (eg. \((xy4)^n\)) have a cyclisity of 2. When n is odd \((xy4)^n\) will end with 4 and when n is even \((xy4)^n\) will end with 6.
• Integers ending with 9 (eg. \((xy9)^n\)) have a cyclisity of 2. When n is odd \((xy9)^n\) will end with 9 and when n is even \((xy9)^n\) will end with 1.