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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

7^1 ends in 7
7^2 ends in 9
7^3 ends in 3
7^4 ends in 1
then it repeates
so, 7^47 will have units digit same as 7^3 --> 3 = a

125^10 * 28^15 = 5^30 * 2^30 * 7^14 = 7*14 * 10^30 which will have 30 zeros preceded by units digit of 7^14 (same as that of 7^3 = 3 = b)

hence, a+b=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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