The last digit in 7 300 is

We have 7 2 = 49 = 50 − 1 Now, 7 300 = (7 2 ) 150 = (50 − 1) 150 $= \ ^{150} C_{0}(50)^{150}(-1)^{0}+^{150} C_{1}(50)^{149}(-1)^{1}+$ $\ldots+^{150} C_{150}(50)^{0}(-1)^{15}$ . Thus the last digits of 7 300 are $^{150} C_{150} \cdot 1.1$ i.e., 1.

Binomial Theorem

Grade 11

Question:

The last digit in 7300 is

9

7

3

1

Step-by-Step Solution

Key Concept: Use the Binomial Theorem to express the power in the form (10n ± 1)^k, where all terms except the last are multiples of 10.

We have 72 = 49 = 50 − 1 Now, 7300 = (72)150 = (50 − 1)150 $= \ ^{150} C_{0}(50)^{150}(-1)^{0}+^{150} C_{1}(50)^{149}(-1)^{1}+$$\ldots+^{150} C_{150}(50)^{0}(-1)^{15}$. Thus the last digits of 7300 are $^{150} C_{150} \cdot 1.1$ i.e., 1.

Correct Answer: D

Question:

Step-by-Step Solution

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