## If $$a$$ is the units digit of $$7^{47}$$ and $$b$$ is the rightmost nonzero digit in $$125^{10}\cdot 28^{15}.$$ What

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### If $$a$$ is the units digit of $$7^{47}$$ and $$b$$ is the rightmost nonzero digit in $$125^{10}\cdot 28^{15}.$$ What

by M7MBA » Thu Sep 17, 2020 1:03 am

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If $$a$$ is the units digit of $$7^{47}$$ and $$b$$ is the rightmost nonzero digit in $$125^{10}\cdot 28^{15}.$$ What is the value of $$a+b?$$

A) 1
B) 2
C) 5
D) 6
F) 8

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### Re: If $$a$$ is the units digit of $$7^{47}$$ and $$b$$ is the rightmost nonzero digit in $$125^{10}\cdot 28^{15}.$$ Wha

by [email protected] » Thu Sep 24, 2020 7:04 am
M7MBA wrote:
Thu Sep 17, 2020 1:03 am
If $$a$$ is the units digit of $$7^{47}$$ and $$b$$ is the rightmost nonzero digit in $$125^{10}\cdot 28^{15}.$$ What is the value of $$a+b?$$

A) 1
B) 2
C) 5
D) 6
F) 8

Solution:

Recall the units digit pattern of powers of 7 is 7-9-3-1. Since the exponent 47 is 3 more than a multiple of 4, 7^47 has a units digit of 3. So a = 3.

Let’s now determine the rightmost nonzero digit of 125^10 x 28^15:

125^10 x 28^15 = (5^3)^10 x (2^2 x 7)^15 = 5^30 x 2^30 x 7^15 = 7^15 x 10^30

We see that the rightmost nonzero digit of 125^10 x 28^15 is the units digit of 7^15 since multiplying by 10^30 means attaching 30 zeros to the end of 7^15.

Since the exponent 15 is also 3 more than a multiple of 4, 7^15 has a units digit of 3. So b = 3. Finally, a + b = 3 + 3 = 6.