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# Reading from the right, how many consecutive zeros, starting with the

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 5600
GMAT 1: 800 Q59 V59
GPA: 3.82
Reading from the right, how many consecutive zeros, starting with the [#permalink]

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09 Mar 2018, 00:23
00:00

Difficulty:

25% (medium)

Question Stats:

74% (00:24) correct 26% (00:51) wrong based on 61 sessions

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[GMAT math practice question]

Reading from the right, how many consecutive zeros, starting with the units digit, appear in 29!?

A. 5
B. 6
C. 7
D. 8
E. 9

_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" examPAL Representative Joined: 07 Dec 2017 Posts: 435 Re: Reading from the right, how many consecutive zeros, starting with the [#permalink] ### Show Tags 09 Mar 2018, 02:16 MathRevolution wrote: [GMAT math practice question] Reading from the right, how many consecutive zeros, starting with the units digit, appear in 29!? A. 5 B. 6 C. 7 D. 8 E. 9 The number of zeros at the right end of the number is determined by the number of 10s in its factorization. That is, we can use basic divisibility properties to answer this question, a Logical approach. We'll count the number of 5s in the prime factorization of 29! : 5 - 1 10 - 1 15 - 1 20 - 1 25 - 2 Total of 6 5s and certainly more than 6 2s meaning that we have 6 pairs of 5*2 = 10 Then there are 6 zeros at the right end of 29! (B) is our answer. _________________ David Senior tutor at examPAL Signup for a free GMAT course We won some awards: Join our next webinar (free) Save up to$250 on examPAL packages (special for GMAT Club members)

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Re: Reading from the right, how many consecutive zeros, starting with the [#permalink]

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11 Mar 2018, 03:01
MathRevolution wrote:
[GMAT math practice question]

Reading from the right, how many consecutive zeros, starting with the units digit, appear in 29!?

A. 5
B. 6
C. 7
D. 8
E. 9

A zero will be formed by $$2*5$$. So we need to find out number of 5s in $$29!$$. this can be calculated as

$$\frac{29}{5}+\frac{29}{5^2}=5+1=6$$

Option B
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 5600
GMAT 1: 800 Q59 V59
GPA: 3.82
Re: Reading from the right, how many consecutive zeros, starting with the [#permalink]

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11 Mar 2018, 18:34
=>

To find the number of $$0’s$$ ending $$29!$$, we need to count the numbers of $$2’s$$ and $$5’s$$ in the prime factorization of $$29!$$.
Since the number $$2’s$$ is greater than the number of $$5’s$$, we only need to count the number of $$5’s$$ in the prime factorization of $$29!$$.
The factors of $$5$$ are contributed by $$5, 10, 15, 20$$ and $$25$$. Each of $$5, 10, 15$$ and $$20$$ contributes one $$5$$, while $$25$$ contributes two $$5s$$ to the prime factorization.
Thus, there are $$6$$ copies of $$5$$ in the prime factorization of $$29!$$, giving rise to $$6$$ consecutive $$0’s$$ at the end of $$29!$$.

Note: The actual value of $$29!$$ is $$8841761993739701954543616000000.$$
_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only \$99 for 3 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"

Re: Reading from the right, how many consecutive zeros, starting with the   [#permalink] 11 Mar 2018, 18:34
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