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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>?
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| − | <cmath>\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\textbf{(D)}~17\qquad\textbf{(E)}~18\qquad</cmath>
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| − | ==Solution 1==
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| − | Prime factorizing this gives us <math>2^{15}\cdot3^{5}\cdot5^{15}</math> Pairing <math>2^{15}</math> and <math>5^{15}</math> gives us a number with <math>15</math> zeros giving us 15 digits. <math>3^5=243</math> and this adds an extra 3 digits. <math>15+3=\text{\boxed{(E)18}}</math>
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| − | ~zhenghua
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| − | ==See Also==
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| − | {{AMC12 box|year=2023|ab=A|num-b=3|num-a=5}}
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| − | {{AMC10 box|year=2023|ab=A|num-b=4|num-a=6}}
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| − | {{MAA Notice}}
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