{"id":122,"date":"2020-01-31T05:30:26","date_gmt":"2020-01-31T05:30:26","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=122"},"modified":"2021-08-03T21:59:09","modified_gmt":"2021-08-03T21:59:09","slug":"mathcount-exercises-19","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=122","title":{"rendered":"MATHCOUNTS Exercises – 19"},"content":{"rendered":"\nFor the following equation $$ x\\times y\\times z = 360$$\n
    \n
  1. Determine the number of unique positive integer solutions.<\/li>\n
  2. Determine the number of unique integer solutions.<\/li>\n<\/ol>\n

    Click here for the solutions.<\/a><\/p>\n

    \n

    Solution for Question 1<\/strong> Because the product of $x$, $y$ and $z$ is $360$, each of them must be a factor of $360$. Because $360=2^3 \\times 3^2 \\times 5$, $360$ has 3 prime factors, $2$, $3$ and $5$. Therefore, each of $x$, $y$ and $z$ can be expressed in the form of $$2^a \\times 3^b \\times 5^c$$ where $a$, $b$ and $c$ are non-negative integers. Let $$ \\begin{array}{rcl}\nx& = & 2^{x_2} \\times 3^{x_3} \\times 5^{x_5} \\\\\ny& = & 2^{y_2} \\times 3^{y_3} \\times 5^{y_5} \\\\\nz& = & 2^{z_2} \\times 3^{z_3} \\times 5^{z_5} \\\\\n\\end{array} $$\nWe have\n$$ x \\times y \\times z = 2^{x_2 + y_2 + z_2}\\times 3^{x_3 + y_3 + z_3}\\times 5^{x_5 + y_5 + z_5} = 2^3 \\times 3^2 \\times 5^1 $$\nTherefore, we have $$ \\begin{array}{cccccr}\nx_2 & + & y_2 & + & z_2 & = & 3 \\\\\nx_3 & + & y_3 & + & z_3 & = & 2 \\\\\nx_5 & + & y_5 & + & z_5 & = & 1 \\\\\n\\end{array} $$\nThe numbers of non-negative integer solutions for the above 3 equations are $$\\begin{array}{ccccc}\n{3 + 3 – 1 \\choose 3 – 1} & = & {5 \\choose 2} & = & 10 \\\\\n{2 + 3 – 1 \\choose 3 – 1} & = & {4 \\choose 2} & = & 6 \\\\\n{1 + 3 – 1 \\choose 3 – 1} & = & {3 \\choose 2} & = & 3 \\\\\n\\end{array} $$\nSo the answer to the question is $ 10\\times 6\\times 3 = 180 $\n<\/p>\n

    Solution for Question 2<\/strong> Because $x$, $y$ and $z$ can be negative integers, either all of them must be positive, or 2 of them must be negative. So the answer to the question is\n$$180 + 180 \\times {3 \\choose 2} = 180 + 180\\times 3 = 720$$<\/p>\n<\/div>\n\n\n\n

    <\/p>\n","protected":false},"excerpt":{"rendered":"

    For the following equation $$ x\\times y\\times z = 360$$ Determine the number of unique positive integer solutions. Determine the number of unique integer solutions. Click here for the solutions. Solution for Question 1 Because the product of $x$, $y$ … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false},"categories":[10,11],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/122"}],"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=122"}],"version-history":[{"count":4,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/122\/revisions"}],"predecessor-version":[{"id":2180,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/122\/revisions\/2180"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=122"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=122"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=122"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}