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If 5^{m-2}2^{m+2} is an n-digit integer, where m is an integer greater

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If 5^{m-2}2^{m+2} is an n-digit integer, where m is an integer greater  [#permalink]

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New post 10 Sep 2018, 01:59
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[Math Revolution GMAT math practice question]

If \(5^{m-2}2^{m+2}\) is an \(n\)-digit integer, where \(m\) is an integer greater than \(2\), what is the value of \(n\), in terms of \(m\)?

\(A. n=m-4\)
\(B. n=m-2\)
\(C. n=m-1\)
\(D. n=m\)
\(E. n=2m\)

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Re: If 5^{m-2}2^{m+2} is an n-digit integer, where m is an integer greater  [#permalink]

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New post 10 Sep 2018, 02:14
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Since M is greater than 2. Lets say M as 3, then

\(5^{3-2}\)\(2^5\) = 32*5 = 160 = 3 digit integer.

Now, Let M be 4. Then

\(5^{4-2}\)\(2^6\) = 1600 = 4 digit number.

Let M be 6. Then

\(5^{6-2}\)\(2^8\) = 160000 = 6 digit number.

We can conclude that n = m.

D is the answer.
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Re: If 5^{m-2}2^{m+2} is an n-digit integer, where m is an integer greater  [#permalink]

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New post 10 Sep 2018, 06:09
Afc0892 wrote:
Since M is greater than 2. Lets say M as 3, then

\(5^{3-2}\)\(2^5\) = 32*5 = 160 = 3 digit integer.

Now, Let M be 4. Then

\(5^{4-2}\)\(2^6\) = 1600 = 4 digit number.

Let M be 6. Then

\(5^{6-2}\)\(2^8\) = 160000 = 6 digit number.

We can conclude that n = m.

D is the answer.


I also think plugging in values is the fastest way! But I think one could stop after one value for m, since it has to be true for all values!
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Re: If 5^{m-2}2^{m+2} is an n-digit integer, where m is an integer greater  [#permalink]

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New post 10 Sep 2018, 06:13
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T1101 wrote:
I also think plugging in values is the fastest way! But I think one could stop after one value for m, since it has to be true for all values!


Yes, But It's always better to check with one or two values just to be on the safer side.
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Re: If 5^{m-2}2^{m+2} is an n-digit integer, where m is an integer greater  [#permalink]

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New post 10 Sep 2018, 07:02
MathRevolution wrote:
[Math Revolution GMAT math practice question]

If \(5^{m-2}2^{m+2}\) is an \(n\)-digit integer, where \(m\) is an integer greater than \(2\), what is the value of \(n\), in terms of \(m\)?

\(A. n=m-4\)
\(B. n=m-2\)
\(C. n=m-1\)
\(D. n=m\)
\(E. n=2m\)

\(5^{m-2}2^{m+2}\)

\(= \frac{5^m*2^m*2^2}{5^2}\)

Now, here if \(m = 3\)

\(= \frac{5^3*2^3*2^2}{5^2}\)

\(= 5*2^5\)

\(= 160\) { 3 digit integer }

Try with \(m = 4\)

So, \(\frac{5^4*2^4*2^2}{5^2}\)

Or, \(5^2*2^6 = 1600\) { 4 digit integer }

Hence, Correct Answer must be (D) \(m = n\)
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Re: If 5^{m-2}2^{m+2} is an n-digit integer, where m is an integer greater  [#permalink]

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New post 10 Sep 2018, 12:00
MathRevolution wrote:
[Math Revolution GMAT math practice question]

If \(5^{m-2}2^{m+2}\) is an \(n\)-digit integer, where \(m\) is an integer greater than \(2\), what is the value of \(n\), in terms of \(m\)?

\(A. n=m-4\)
\(B. n=m-2\)
\(C. n=m-1\)
\(D. n=m\)
\(E. n=2m\)


I AGREE that exploring a particular case is THE way to go in this problem. And, yes, m=3 is great!

I offer below the solution for the general case, (I hope) interesting in terms of "using the exercise as a good way to strengthen our skills"...

\({5^{\,m - 2}}\, \cdot {2^{\,m + 2}} = \boxed{\,n\,}{\text{ - digit}}\,\,\operatorname{int} \,\,\,\,\,\,\left( {m \geqslant 3\,\,\,\operatorname{int} } \right)\)

\(?\,\,\,:\,\,\,n = f\left( m \right)\)


\({5^{\,m - 2}}\, \cdot {2^{\,m + 2}} = {5^{\,m - 2}}\, \cdot {2^{\,m - 2}} \cdot {2^{\,4}} = 32 \cdot {10^{\,m - 2}}\,\,\,\,\,\,\left[ {m - 2 \geqslant 1\,\,\,\operatorname{int} } \right]\)


Now let´s look for a PATTERN:

\(32 \cdot {10^{\,\boxed3 - 2}} = 320\,\,\,\,\, \to \,\,\,\,\,\boxed3\,\,\,{\text{digits}}\,\)

\(32 \cdot {10^{\,\boxed4 - 2}} = 3200\,\,\,\,\, \to \,\,\,\,\,\boxed4\,\,\,{\text{digits}}\,\)

\(\ldots\)

\(32 \cdot {10^{\,\boxed{\,n\,} - 2}} = 320 \ldots 0\,\,\,\,\, \to \,\,\,\,\,\boxed{\,n\,}\,\,\,{\text{digits}}\)


And we are done!

If you prefer, consider one more line of argument:

\(32 \cdot {10^{\,m - 2}}\,\,\,\, = \,\,\,\,32 \cdot {10^{\,n - 2}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,m = n\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.
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Re: If 5^{m-2}2^{m+2} is an n-digit integer, where m is an integer greater  [#permalink]

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New post 12 Sep 2018, 02:42
=>

\(5^{m-2}2^{m+2} = 5^{m-2}2^{m-2+4} = 5^{m-2}2^{m-2}*2^4 = (5*2)^{m-2}*2^4= (10)^{m-2}*2^4\)
\(= 16*(10)^{m-2} = 16*(100…0)= 1600…0\) with \(m-2\) 0’s.
\(10^{m-2}\) has \(m-2\) digits that are \(0\) and \(16*(10)^{m-2}\) has digits including \(1, 6\) and \(m-2\) digits that are \(0\).
Thus, \(16*(10)^{m-2}\) has \(m – 2 + m = m\) digits.

Therefore, D is the answer.
Answer: D
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Re: If 5^{m-2}2^{m+2} is an n-digit integer, where m is an integer greater   [#permalink] 12 Sep 2018, 02:42
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