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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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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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New post 01 May 2018, 05:55
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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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New post 01 May 2018, 06:14
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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.

Answer: B
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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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New post 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.

Answer: B
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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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New post 22 Jan 2019, 16:06
1
My answer:

Idk if my approach is correct, could someone please help me to verify it?

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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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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New post 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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New post 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
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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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New post 21 Jan 2019, 22:58
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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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New post 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

Please advise if this logic is flawed.
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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] 25 Jan 2019, 14:33
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