Difference between revisions of "2023 AMC 10A Problems/Problem 5"

Latest revision as of 23:15, 9 November 2023

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-==Problem==
+#redirect[[2023 AMC 12A Problems/Problem 4]]
-How many digits are in the base-ten representation of <math>8^5 \cdot 5^{10} \cdot 15^5</math>?
-<math>\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\textbf{(D)}~17\qquad\textbf{(E)}~18\qquad</math>
-==Solution 1==
-Prime factorization of this gives us <math>2^{15}\cdot3^{5}\cdot5^{15}</math>. Pairing <math>2^{15}</math> and <math>5^{15}</math> together gives us a number with <math>15</math> zeros, or 15 digits. <math>3^5=243</math> and this adds an extra 3 digits. <math>15+3=\boxed{\textbf{(E) 18}}</math>
-~zhenghua ~not_slay
-==See Also==
-{{AMC12 box|year=2023|ab=A|num-b=3|num-a=5}}
-{{AMC10 box|year=2023|ab=A|num-b=4|num-a=6}}
-{{MAA Notice}}