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# The quantity 3^3 4^4 5^5 6^6 - 3^6 4^5 5^4 6^3 will end in how many

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Math Expert
Joined: 02 Sep 2009
Posts: 56319
The quantity 3^3 4^4 5^5 6^6 - 3^6 4^5 5^4 6^3 will end in how many  [#permalink]

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01 May 2018, 05:55
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Question Stats:

67% (02:03) correct 33% (02:08) wrong based on 181 sessions

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The quantity $$3^3 4^4 5^5 6^6 - 3^6 4^5 5^4 6^3$$ will end in how many zeros ?

A. 3

B. 4

C. 5

D. 6

E. 9

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Posts: 142
Location: Uzbekistan
Re: The quantity 3^3 4^4 5^5 6^6 - 3^6 4^5 5^4 6^3 will end in how many  [#permalink]

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01 May 2018, 06:14
1
1
Bunuel wrote:
The quantity $$3^3 4^4 5^5 6^6 - 3^6 4^5 5^4 6^3$$ will end in how many zeros ?

A. 3

B. 4

C. 5

D. 6

E. 9

Let's rewrite the expression as follows:
$$3^3*4^4*5^5*6^6 - 3^6*4^5*5^4*6^3 = 3^3*4^4*5^4*6^3*(5*6^3 - 3^3*4)$$

$$3^3*4^4*5^4*6^3$$ will end in $$4$$ zeros.
$$(5*6^3 - 3^3*4)$$ will not end in zero (it will end in $$2$$).

Hence, the expression ends in $$4$$ zeros.

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Re: The quantity 3^3 4^4 5^5 6^6 - 3^6 4^5 5^4 6^3 will end in how many  [#permalink]

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02 May 2018, 10:22
1
Bunuel wrote:
The quantity $$3^3 4^4 5^5 6^6 - 3^6 4^5 5^4 6^3$$ will end in how many zeros ?

A. 3

B. 4

C. 5

D. 6

E. 9

Simplifying into primes we have:

3^3 x 2^8 x 5^5 x 2^6 x 3^6 - 3^6 x 2^10 x 5^4 x 2^3 x 3^3

3^9 x 2^14 x 5^5 - 3^9 x 2^13 x 5^4

Factoring out we have:

3^9 x 2^13 x 5^4(1 x 2 x 5 - (1 x 1 x 1))

3^9 x 2^13 x 5^4 x 9

We know that each occurrence of 10 in a factorization yields one trailing zero. Note that a “5 and 2” pair in a factorization is equivalent to a 10. Since we have four “5 and 2” pairs,, we have 4 trailing zeros.

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Re: The quantity 3^3 4^4 5^5 6^6 - 3^6 4^5 5^4 6^3 will end in how many  [#permalink]

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12 Jan 2019, 09:11
Can we not simply ignore every number that does not produce a trailing zero: $$5^5-5^4=5^4(5-1)$$ $$\implies$$ 4 trailing zeros?
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The quantity 3^3 4^4 5^5 6^6 - 3^6 4^5 5^4 6^3 will end in how many  [#permalink]

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21 Jan 2019, 22:19
Bunuel wrote:
The quantity $$3^3 4^4 5^5 6^6 - 3^6 4^5 5^4 6^3$$ will end in how many zeros ?

A. 3

B. 4

C. 5

D. 6

E. 9

My reasoning if it helps anyone:

Break everything into primes, then factor out as much as you can to get rid of the subtraction element of this question.

$$3^9 * (2^{14}) * 5^5 - 3^9 * (2^{13}) * 5^4$$

$$3^9 * (2^{13}) * 5^4 (2*5 - 1)$$

The number of zeros a number will have is determined by how many times you multiply 10 to it.

In the above we can multiply 5*2 to get 10. We have four 5's, so we can create four 10's, hence our answer will have 4 zeros.

EDIT: it's supposed to say 2^14 and 2^13. I'm not sure why the math tag isn't working
Math Expert
Joined: 02 Sep 2009
Posts: 56319
Re: The quantity 3^3 4^4 5^5 6^6 - 3^6 4^5 5^4 6^3 will end in how many  [#permalink]

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21 Jan 2019, 22:58
kchen1994 wrote:
EDIT: it's supposed to say 2^14 and 2^13. I'm not sure why the math tag isn't working

When you have more than one character in exponents, put it in { }: 2^{123} --> $$2^{123}$$
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Re: The quantity 3^3 4^4 5^5 6^6 - 3^6 4^5 5^4 6^3 will end in how many  [#permalink]

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22 Jan 2019, 16:06
1

As we are searching for the number of 0's then we just have to look for the 2 and 5 pairs.

The limiting factor will be the 5's.

Five 0's - Four 0's

100000 - 10000 = 99..0000.

Hence... B
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Joined: 05 Jan 2019
Posts: 10
Re: The quantity 3^3 4^4 5^5 6^6 - 3^6 4^5 5^4 6^3 will end in how many  [#permalink]

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25 Jan 2019, 14:33
Hello,

How I approached answering this question :

To find: Number of trailing 0's after performing subtraction

Approach: I only looked at 2's and 5's on both the sides

4^4 * 5^5 = 2^8 * 5^5 {this will give me 5 trailing 0's at the end (2^5 * 5^5 - need to consider highest power of 5) }
Similarly, 4^5*5*4 = 2^10*5^4 { this will give me 4 trailing 0's at the end (2^4*5^4)}
so, now I have 100000-10000, this gives me 4 trailing 0's at the end.

Ans: B

Re: The quantity 3^3 4^4 5^5 6^6 - 3^6 4^5 5^4 6^3 will end in how many   [#permalink] 25 Jan 2019, 14:33
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