Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 28"

Latest revision as of 16:13, 6 May 2007

The product of $15^8\cdot28^6\cdot5^{11}$ is an integer number whose last digits are zeros. How many zeros are there?

$\mathrm{(A) \ } 6\qquad \mathrm{(B) \ } 8\qquad \mathrm{(C) \ } 11\qquad \mathrm{(D) \ } 12\qquad \mathrm{(E) \ } 19$

The number of zeros at the end of a number is determined by the number of 2's and 5's in its prime factorization.

$15^8\cdot28^6\cdot5^{11}=2^{12}\cdot3^8\cdot5^{19}\cdot7^6$

There are 12 $2$ 's and 19 $5$ 's. Each pair adds a zero, but any extras don't count (in this case, the 7 extra $5$ 's).

$12\Longrightarrow\mathrm{ D}$

@@ Line 5: / Line 5: @@
 ==Solution==
-The number of zeros at the end of a number is determined by the number of 2's and 5's in its prime factorization.
+The number of zeros at the end of a number is determined by the number of 2's and 5's in its [[prime factorization]].
 <math>15^8\cdot28^6\cdot5^{11}=2^{12}\cdot3^8\cdot5^{19}\cdot7^6</math>
-There are 12 2's and 19 5's. Each pair adds a zero, but any extras don't count (in this case, the 7 extra 5's).
+There are 12 <math>2</math>'s and 19 <math>5</math>'s. Each pair adds a zero, but any extras don't count (in this case, the 7 extra <math>5</math>'s).
-<math>12\Rightarrow\mathrm{ D}</math>
+<math>12\Longrightarrow\mathrm{ D}</math>
 ==See also==
-*[[2007 Cyprus MO/Lyceum/Problems]]
+{{CYMO box|year=2007|l=Lyceum|num-b=27|num-a=29}}
-*[[2007 Cyprus MO/Lyceum/Problem 27|Previous Problem]]
+[[Category:Introductory Algebra Problems]]
-*[[2007 Cyprus MO/Lyceum/Problem 29|Next Problem]]