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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, 04:55
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

70% (02:58) correct 30% (01:38) wrong based on 91 sessions

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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, 05:14
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.

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, 09:22
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 12 Jan 2019, 08:11
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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Re: The quantity 3^3 4^4 5^5 6^6 - 3^6 4^5 5^4 6^3 will end in how many &nbs [#permalink] 12 Jan 2019, 08:11
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