{"id":3483,"date":"2023-01-13T07:29:00","date_gmt":"2023-01-13T11:29:00","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=3483"},"modified":"2024-10-25T10:25:29","modified_gmt":"2024-10-25T14:25:29","slug":"aime-2017-i-problem-14","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=3483","title":{"rendered":"AIME 2017 I – Problem 14"},"content":{"rendered":"\n

Let $a > 1$ and $x > 1$ satisfy $$\\log_a(\\log_a(\\log_a 2) + \\log_a 24 – 128) = 128$$ and $\\log_a(\\log_a x) = 256$. Find the remainder when $x$ is divided by $1000$. \ud83d\udd11<\/a><\/p>\n\n\n\n

\n\n\n\n

Solution: <\/strong>Let $a=2^n$, we have $$log_{2^n}(\\log_{2^n}(\\log_{2^n}2)+\\log_{2^n}24-128)=128$$<\/p>\n\n\n\n

$$log_{2^n}(\\log_{2^n}2)+\\log_{2^n}24-128=2^{128n}$$<\/p>\n\n\n\n

$$log_{2^n}(\\log_{2^n}2^{24})=2^{128n}+128$$<\/p>\n\n\n\n

$$log_{2^n}(\\dfrac{24}{n})=2^{128n}+128$$<\/p>\n\n\n\n

$$\\dfrac{1}{n}log_2(\\dfrac{24}{n})=2^{128n}+128$$<\/p>\n\n\n\n

$$log_2(\\dfrac{24}{n})=n(2^{128n}+128)$$<\/p>\n\n\n\n

Let $n=3\\cdot 2^m$, we have $$3-m=3\\cdot 2^m(2^{384\\cdot 2^m}+128)$$<\/p>\n\n\n\n

Check the range of $m$, we found $-7<m<-5$ and $m=-6$ is a solution. Therefore $n=3\\cdot 2^{-6}=\\dfrac{3}{64}$, and $a=2^{\\frac{3}{64}}$. As $log_ax=a^{256}$, we have<\/p>\n\n\n\n

$$log_{2^{\\frac{3}{64}}}x=2^{256\\cdot{\\frac{3}{64}}}=2^{12}$$<\/p>\n\n\n\n

$$\\dfrac{64}{3}log_{2}x=2^{12}$$<\/p>\n\n\n\n

$$log_{2}x=192$$<\/p>\n\n\n\n

$$x=2^{192}$$<\/p>\n\n\n\n

$$x|1000=2^{192}|1000=4\\cdot 2^{190}|1000=4\\cdot 1024^{19}|1000=4\\cdot 24^{19}|1000$$<\/p>\n\n\n\n

$$=96\\cdot 24^{18}|1000=96\\cdot 13824^6|1000=96\\cdot 176^6|1000=96\\cdot 30976^3|1000$$<\/p>\n\n\n\n

$$=-96\\cdot 24^3|1000=-96\\cdot 13824|1000=96\\cdot 176|1000=16896|1000=\\boxed{896}$$ <\/p>\n\n\n\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Let $a > 1$ and $x > 1$ satisfy $$\\log_a(\\log_a(\\log_a 2) + \\log_a 24 – 128) = 128$$ and $\\log_a(\\log_a x) = 256$. Find the remainder when $x$ is divided by $1000$. \ud83d\udd11 Solution: Let $a=2^n$, we have $$log_{2^n}(\\log_{2^n}(\\log_{2^n}2)+\\log_{2^n}24-128)=128$$ $$log_{2^n}(\\log_{2^n}2)+\\log_{2^n}24-128=2^{128n}$$ … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false},"categories":[13],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/3483"}],"collection":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=3483"}],"version-history":[{"count":25,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/3483\/revisions"}],"predecessor-version":[{"id":4594,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/3483\/revisions\/4594"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=3483"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=3483"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=3483"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}