Question

last digit of (4^n) + (7^n) when n is equal to 97 is ?​

Answer 1

Answer:

197

Step-by-step explanation:

Maybe

Answer 2

The last digit of 4ⁿ + 7ⁿ when n is equal to 97 is 1.

Given

• 4ⁿ + 7ⁿ
• n = 97

To Find

The last digit of the number

Solution

If we observe the pattern for powers of 4 here, we will see that

4¹ = 4

4² = 16

4³ = 64

4⁴ = 256

4⁵ = 1024

4⁶ = 4096

So we can observe that the last digits of powers of 4 follow the pattern of 4, 6, 4, 6, 4,...

Thus-

All odd powers of 4 have last digits as 4 and all even of 4 powers have their last digits as 6

Therefore the last digit of 4ⁿ where n = 97 is 4

Similarly, if we observe the powers of 7

7¹ = 7

7² = 49

7³ = 343

7⁴ = 2401

7⁵ = 16807

7⁶ = 117649

7⁷ = 823543

7⁸ = 5764801

here, the last digits repeat in 4s i.e 7, 9, 3, 1

97/4 = $24\frac{1}{4}$

Therefore, for 7⁹⁷, if we start with 7, 7²,... the complete cycle will be repeated 24 times, and that one extra multiplication will give us the last digit 7

Therefore the last digit of 7ⁿ where n = 97 is 7

Adding these 2 digits gives us 11.

Now the 1 in tens place will be carried over, to give us our last digit 1 for 4ⁿ + 7ⁿ when n is equal to 97.

Hence the last digit of 4ⁿ + 7ⁿ when n is equal to 97 is 1.

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