The bad news is you can't use a calculator on GMAT. The good news is most calculators won't show this answer - they will truncate and use scientific notation. Why is that good news? GMAT doesn't expect you to be a human calculator.


Identify the problem. It's an exponent problem and a product problem.


Set up the problem. Write what you know about the type/subject, and write the info the problem gives.

What do we know about exponents? They follow patterns.

Whenever there's a seemingly impossibly long pattern on the GMAT, check the first few numbers in the pattern. You'll find it isn't impossible.

The final digit of a number raised to some exponent can be, at most, one of 0,1,2,3,4,5,6,7,8,9 - that's a finite pattern with only 10 possibilities.

What do we know about products? The final digit of a product of two numbers is the final digit of the product of the final digits of each number.

348 x 976 has a final digit 8. That's the final digit of 8x6 = 48. So all we have to worry about here is the final digits of each number. And each "impossibly" large number has only 10 possibilities for its final digit. This isn't really hard.

What does the problem tell us?

7^95 - check the last digit pattern:

7^1 = 7
7^2 = 49
7^3 = 343
7^4 = 2401
7^5 = 16807 - the pattern already repeats. It repeats every 5th number exponent. So divide the exponent by 4. If the remainder is:
0 - last digit 1
1 - last digit 7
2 - last digit 9
3 - last digit 3

95 / 4 = 23 remainder 3. The last digit is 3


3^58 - check the last digit pattern.

3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243 It repeats every 5th number exponent. So divide the exponent by 4. If the remainder is:

0 - last digit 1
1 - last digit 3
2 - last digit 9
3 - last digit 7

58 / 4 = 14 R 2. The last digit is 9.


Solve the problem.

7^95 last digit is 3. 3^58 last digit is 9. 3 x 9 = 27. The last digit is 7. That's choice d.

3 x 9 = 27, The last digit is 7.

Keep in mind that all numbers raised to powers have a cycle. If you memorize the cycles, these problems are a breeze.

https://cat.wordpandit.com/concepts/cat- ... -a-number/

Solve the problem.

7^95 last digit is 3. 3^58 last digit is 9. 3 x 9 = 27. The last digit is 7. That's choice d.

3 x 9 = 27, The last digit is 7.

Thanks fcabanski, but why will you multiply ? The question has a sign of subtraction.

Sachin

fcabanski wrote:I multiplied because I wasn't paying attention. Or I multiplied because I was, er, testing you to see if you were paying attention. Yeah, that's it.

The problem is subtraction and exponents.

7^95 - last digit is 3.
3^58 - last digit is 9.

When subtracted it's ...x3 - ...y9 (x and y represent the digits before the 3 and 9). Borrowing 10 from the digit in front of the 3 it becomes:

13 - 9 = 4. That's answer b.