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# What is the rightmost non-zero digit of 30^58 x 17^85?

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What is the rightmost non-zero digit of 30^58 x 17^85? [#permalink]

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30 Jul 2017, 13:08
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What is the rightmost non-zero digit of $$30^{58}*17^{85}$$?

A. 0
B. 1
C. 3
D. 7
E. 9

Thanks,
Saquib
Quant Expert
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Re: What is the rightmost non-zero digit of 30^58 x 17^85? [#permalink]

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30 Jul 2017, 13:09
Reserving this space to post the official solution.
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Re: What is the rightmost non-zero digit of 30^58 x 17^85? [#permalink]

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31 Jul 2017, 01:15
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EgmatQuantExpert wrote:
What is the rightmost non-zero digit of $$30^{58}*17^{85}$$?

A. 0
B. 1
C. 3
D. 7
E. 9

First, need to make a Prime Factorization.

$$30^{58}*17^{85} = (2*3*5)^{58} * 17 ^{85} = (2*5)^{58} * 3^{58} * 17 ^{85}$$

Note that $$(2*5)^{58} = 10^{58}$$ lead to zero digit. We just need to calculate the digit number of $$3^{58} * 17^{85}$$

Note that $$3^4=81=(...1)$$.
Hence we have $$3^{58}=(3^4)^{14}*3^2=(...1)^{14} * 9 = (...9)$$

Note that $$17^2=(...9)$$.
Hence we have $$17^4 = (...9)^2 = (...1) \implies 17^85=(17^4)^{21}*17=(...1)^{21} * 17 = (...7)$$

Hence $$3^{58} * 17^{85} = (...9) * (...7) = (...3)$$

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What is the rightmost non-zero digit of 30^58 x 17^85? [#permalink]

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01 Aug 2017, 16:27
EgmatQuantExpert wrote:
What is the rightmost non-zero digit of $$30^{58}*17^{85}$$?

A. 0
B. 1
C. 3
D. 7
E. 9

Thanks,
Saquib
Quant Expert
e-GMAT

To 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.

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!

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Re: What is the rightmost non-zero digit of 30^58 x 17^85? [#permalink]

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01 Aug 2017, 17:09
EgmatQuantExpert wrote:
What is the rightmost non-zero digit of $$30^{58}*17^{85}$$?

A. 0
B. 1
C. 3
D. 7
E. 9

Let’s simplify 30^58:

30^58 = 3^58 x 10^58

So, we have:

3^58 x 10^58 x 17^85

Since 10^58 is the number 1 followed by 58 zeros, we really need to determine the units digit of 3^58 and 17^85 (or 7^85):

We can evaluate 3^n for positive integer values of n. That is, let’s look at the pattern of the units digits of powers of 3. When writing out the pattern, notice that we are ONLY concerned with the units digit of 3 raised to each power.

3^1 = 3

3^2 = 9

3^3 = 7

3^4 = 1

3^5 =3

The pattern of the units digits of powers of 3 repeats every 4 exponents. The pattern is 3–9–7–1. In this pattern, all positive exponents that are multiples of 4 will have 1 as their units digit. Thus:

Following the pattern, we see that 3^56 has a units digit of 1, 3^57 has a units digit of 3, and 3^58 has a units digit of 9.

Next, we can evaluate 7^n for positive integer values of n. That is, let’s look at the pattern of the units digits of powers of 7. When writing out the pattern, notice that we are ONLY concerned with the units digit of 7 raised to each power.

7^1 = 7

7^2 = 9

7^3 = 3

7^4 = 1

7^5 = 7

The pattern of the units digits of powers of 7 repeats every 4 exponents. The pattern is 7–9–3–1. In this pattern, all positive exponents that are multiples of 4 will have 1 as their units digit. Thus:

7^84 has a units digit of 1 and 7^85 has a units digit of 7.

Since the units digit of 3^58 is 9 and the units digit of 7^85 is 7, and the product of 9 and 7 is 63, we know that the last non-zero digit of 30^58 and 17^85 is 3.

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Re: What is the rightmost non-zero digit of 30^58 x 17^85? [#permalink]

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01 Aug 2017, 22:13
broall wrote:
EgmatQuantExpert wrote:
What is the rightmost non-zero digit of $$30^{58}*17^{85}$$?

A. 0
B. 1
C. 3
D. 7
E. 9

First, need to make a Prime Factorization.

$$30^{58}*17^{85} = (2*3*5)^{58} * 17 ^{85} = (2*5)^{58} * 3^{58} * 17 ^{85}$$

Note that $$(2*5)^{58} = 10^{58}$$ lead to zero digit. We just need to calculate the digit number of $$3^{58} * 17^{85}$$

Note that $$3^4=81=(...1)$$.
Hence we have $$3^{58}=(3^4)^{14}*3^2=(...1)^{14} * 9 = (...9)$$

Note that $$17^2=(...9)$$.
Hence we have $$17^4 = (...9)^2 = (...1) \implies 17^85=(17^4)^{21}*17=(...1)^{21} * 17 = (...7)$$

Hence $$3^{58} * 17^{85} = (...9) * (...7) = (...3)$$

Hi
why I cannot get the same response by simply calculating exponent cycles for primes (2∗3∗5)^58∗17^85
From my understanding you ignored 10^58 just for the sake of calculation simplicity as it gives 0

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Re: What is the rightmost non-zero digit of 30^58 x 17^85? [#permalink]

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13 Sep 2017, 11:58
Just to do calculations faster we can take this approach also

(10^58 ) ( 3^58) (17^85)

There is a formula a^n * b^n = (a*b)^n

so 85 can be written as 58 +27

( 3^58) (17^58) (17^27)

(51)^58 (17)^27

1^58 and 7^27 will give us the non zero digit

1 and 3

=3
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Re: What is the rightmost non-zero digit of 30^58 x 17^85?   [#permalink] 13 Sep 2017, 11:58
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