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

Max@Math Revolution 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

Since m=2 will yield an integer value for the given expression and will make the math easier, ignore the condition that m>2 and let m=2.
Plugging m=2 into 5^(m-2) * 2^(m+2), we get:
5�2� = 1*16 = 16.
Since the result is a 2-digit integer, n=2.
The correct answer must work for n=m=2.
Only D is viable.

The correct answer is D.

Max@Math Revolution 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

Mitch´s idea of exploring a particular case is THE way to go, although I would prefer m=3, only because of the question stem restriction.

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 the 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, just use one more line:
\[32 \cdot {10^{\,m - 2}}\,\,\,\, = \,\,\,\,32 \cdot {10^{\,n - 2}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,m = n\]

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

Regards,
fskilnik.

=>

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