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 and move right until you get to the last ("rightmost") digit that is not zero.

Example: Imagine the number 987,650,000. The rightmost -----> non-zero digit here is [5]. Visually

987,6[5]0,000

------>[5]0,000

Hope it helps!