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Last digit of any power The last digits of the powers of any number follow a cyclic pattern i.e. They repeat after certain number of steps. Take "2" for example Last digit of 2^1 is 2. Last digit of 2^2 is 4. Last digit of 2^3 is 8. Last digit of 2^4 is 6. Last digit of 2^5 is 2. As you can see the last digit of 2^5 is same as that of 2^1, from this onwards the last digit will keep on repeating in a cyclic pattern. Example 1: Suppose we want to calculate the last digit of 2^67, Solution 1: Since 64 is a multiple of 4, the last digit of 2^64 will be same as for 2^4. The last digits of 2^65, 2^66, 2^67 will be same as for the last digits of 2^1, 2^2, 2^3. Hence 2^67 is the same as the last digit of 2^3 i.e. 8. Similarly we can calculate for 3^74 by the cyclic pattern. |
Last digit of a sum or product Example 2: Find out the last digit of (2^67) + (3^74) Solution 2: Individually calculate the last digit of 2^67 and 3^74. As calculated above last digit of 2^67 is 8. Similarly, the last digit of 3^74 comes out to be 9. Hence, the last digit of (2^67) + (3^74) will be last digit of 2^67 + last digit of 3^74. I.e. 8+9=17.Hence last digit is 7. Example 3: Find out the last digit of (2^67) * (3^74) Solution 3: Individually calculate the last digit of 2^67 and 3^74. As calculated above last digit of 2^67 is 8. Similarly, the last digit of 3^74 comes out to be 9. Hence, the last digit of (2^67) * (3^74) will be last digit of 2^67 * last digit of 3^74. I.e. 8*9=72.Hence last digit is 2. |
