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.
_________________
Fabio Skilnik ::
GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here:
https://gmath.net