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 :: https://www.GMATH.net (Math for the GMAT)

Course release PROMO : finish our test drive till 30/Nov with (at least) 50 correct answers out of 92 (12-questions Mock included) to gain a 50% discount!