e-GMAT Question:

If \(a\) is the units digit of \(7^{47}\) and b is the rightmost nonzero digit in \((125^{10}× 28^{15})\). What is the value of \(a+b\)?
A) 1
B) 2
C) 5
D) 6
F) 8

This is

Question 6 of The e-GMAT Number Properties Marathon

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Question 7 of the Marathon

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The cyclicity of number 2 and 7 will be needed to solve the question

-n--------1---2---3---4
-2--------2---4---8---6
-7--------7---9---3---1

Since a is the units digit of \(7^{47}\), it will be the units digit of \(7^{4*11 + 3}\) or \(7^3\), which is 3.

Similarly, b is the units digit of \((125^{10}× 28^{15})\)

\((125^{10} * 28^{15})\) = \((5^3)^{10} * (2^2 * 7)^{15}\) = \(5^{30}*2^{30}*7^{4*3 + 3}\) = \(10^{30} * 7^{4*3 + 3}\) as \(a^m * b^ m = (ab)^m\)

This will translate to the units digit of \(7^3\) which is 3.

Therefore, the value of a+b is 6(Option D)
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Solution:

We need to the find the sum of \(a\) and \(b\) where, \(a\) is the units digit of \(7^{47}\) and \(b\) is the rightmost nonzero digit in \(125^{10}× 28^{15}\).
Let’s now calculate \(a\)and \(b\).

Units digit of\(7^{47}\)

\(47= 4*11+3\)
\(4*11+3= 4k+3\)
Units digit of \(7^(4k+3)\) is \(3\)

.
Thus, units digit of \(7^{47}\) is \(3\).

\(a= 3\)

Rightmost nonzero digit in 125^{10}× 28^{15}

125^{10}× 28^{15}= (5^3)^10× (2^2*7)^15.
125^{10}× 28^{15}= 5^30 ×2^30 ×7^15
125^{10}× 28^{15}= 10^30 ×7^15

We know, 10^n only gives n number of zeroes at the end of the number.
Hence, the units digit of 7^15 will be same as the rightmost nonzero digit of 125^{10}× 28^{15}.

\(15= 4*3+3\)
\(4*3+3= 4k+3\)
Units digit of \(7^(4k+3)\) is \(3\).

Thus,

\(b= 3\)

Hence,

\(a+b= 3+3\)
\(a+b=6\)

Answer: Option D
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