EgmatQuantExpert wrote:
What is the rightmost non-zero digit of \(30^{58}*17^{85}\)?
Thanks,
Saquib
Quant Expert
e-GMATTo find the rightmost non-zero digit, first get rid of zeros by factoring out any powers of 10.
Here the powers of 10 are in \(30^{58}\), which = \(3^{58}\) * \(10^{58}\). Factor out \(10^{58}\). The whole expression is now \(10^{58}\) * \((3^{58}*17^{85})\)
Ignore the powers of 10. Finding units digit involves only the units digit of the two numbers in parentheses: \(3^{58}\) and \(7^{85}\)
Units digit for both is determined by "cyclicity" of powers of 3 and 7:
\(3^1\) = 3 | \(7^1\) = 7
\(3^2\) = 9 | \(7^2\) = _9
\(3^3\) = _7| \(7^3\) = _3
\(3^4\) = _1| \(7^4\) = _1
---------------------------
\(3^5\) = _3| \(7^5\) = _7
\(3^6\) = _9| \(7^6\) = _9
After four powers, 3 and 7 repeat their patterns. Cyclicity is 4.
Units digit of \(3^{58}\)? Divide power by 4. (58/4) leaves a remainder of
2. The remainder determines where the 58th power "falls" in the cycle. Any power that leaves remainder of
2 has the same units digit as 3 to the power of
2. Units digit for \(3^{58}\) is same as \(3^2\), which is 9.
\(7^{85}\): 85/4 leaves remainder of 1. R1 has same units digit as \(7^1\), which is 7.
Finally, take the two units digits and multiply them together: 9*7 = 63. Rightmost non-zero digit is 3.
ANSWER C
Responding to an email: "rightmost" is literally the NON-zero digit "most to the right." Start from the first digit of the number and move right until you get to the last ("rightmost") digit that is not zero.
Example: Imagine the number 987,650,000.
The first digit is 9. Move to the right-->
The rightmost -----> non-zero digit here is [5]. Visually
987,6[5]0,000
------>[5]0,000
Hope it helps!
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