{"id":25970,"date":"2025-06-19T08:42:03","date_gmt":"2025-06-19T08:42:03","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=25970"},"modified":"2025-06-19T08:42:06","modified_gmt":"2025-06-19T08:42:06","slug":"find-the-smallest-number-by-which-132-must-be-multiplied-so-that-the-resulting-number-is-a-perfect-cube","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/find-the-smallest-number-by-which-132-must-be-multiplied-so-that-the-resulting-number-is-a-perfect-cube\/","title":{"rendered":"find the smallest number by which 132 must be multiplied so that the resulting number is a perfect cube"},"content":{"rendered":"\n

find the smallest number by which 132 must be multiplied so that the resulting number is a perfect cube<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To find the smallest number by which 132 must be multiplied so that the resulting number becomes a perfect cube<\/strong>, we need to analyze the prime factorization<\/strong> of 132.<\/p>\n\n\n\n

\n\n\n\n

Step 1: Prime Factorization of 132<\/h3>\n\n\n\n

132 \u00f7 2 = 66
66 \u00f7 2 = 33
33 \u00f7 3 = 11
11 is a prime number.<\/p>\n\n\n\n

So,
132 = 2 \u00d7 2 \u00d7 3 \u00d7 11 = 2\u00b2 \u00d7 3 \u00d7 11<\/strong><\/p>\n\n\n\n

\n\n\n\n

Step 2: Check for Cube Requirements<\/h3>\n\n\n\n

For a number to be a perfect cube<\/strong>, all exponents<\/strong> in its prime factorization must be multiples of 3<\/strong>.<\/p>\n\n\n\n

Currently:<\/p>\n\n\n\n

    \n
  • 2 appears with exponent 2<\/strong><\/li>\n\n\n\n
  • 3 appears with exponent 1<\/strong><\/li>\n\n\n\n
  • 11 appears with exponent 1<\/strong><\/li>\n<\/ul>\n\n\n\n

    To make it a cube:<\/p>\n\n\n\n

      \n
    • We need 1 more 2<\/strong> (to make 2\u00b3)<\/li>\n\n\n\n
    • We need 2 more 3s<\/strong> (to make 3\u00b3)<\/li>\n\n\n\n
    • We need 2 more 11s<\/strong> (to make 11\u00b3)<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      Step 3: Find the Least Number to Multiply<\/h3>\n\n\n\n

      We need to multiply 132 by the missing factors:
      2 \u00d7 3\u00b2 \u00d7 11\u00b2 = 2 \u00d7 9 \u00d7 121 = 2178<\/strong><\/p>\n\n\n\n

      So,
      132 \u00d7 2178 = a perfect cube<\/strong><\/p>\n\n\n\n

      \n\n\n\n

      \u2705 Final Answer: 2178<\/strong><\/h3>\n\n\n\n
      \n\n\n\n

      \u270d\ufe0f Explanation<\/h3>\n\n\n\n

      To transform a number into a perfect cube, all the prime factors<\/strong> in its factorization must have exponents divisible by 3<\/strong>. This is because a perfect cube is a number that can be written in the form x3x^3×3, which means every prime factor of x3x^3×3 is raised to a power that is a multiple of 3.<\/p>\n\n\n\n

      We begin by factoring 132 into primes:
      132 = 2\u00b2 \u00d7 3 \u00d7 11<\/strong>.<\/p>\n\n\n\n

      Now, we examine each exponent:<\/p>\n\n\n\n

        \n
      • The factor 2 appears twice (2\u00b2), but we need it to appear three times (2\u00b3), so we are missing one 2<\/strong>.<\/li>\n\n\n\n
      • The factor 3 appears once (3\u00b9), but we need it to appear three times (3\u00b3), so we are missing two 3s<\/strong>.<\/li>\n\n\n\n
      • The factor 11 appears once (11\u00b9), but we need it to appear three times (11\u00b3), so we are missing two 11s<\/strong>.<\/li>\n<\/ul>\n\n\n\n

        To compensate for the missing primes, we multiply 132 by:
        2 \u00d7 3 \u00d7 3 \u00d7 11 \u00d7 11 = 2 \u00d7 9 \u00d7 121 = 2178<\/strong><\/p>\n\n\n\n

        When we multiply 132 by 2178, the resulting number has a prime factorization of:
        2\u00b3 \u00d7 3\u00b3 \u00d7 11\u00b3<\/strong>, which is a perfect cube.<\/p>\n\n\n\n

        Therefore, the smallest number<\/strong> by which 132 must be multiplied to get a perfect cube is 2178<\/strong>.<\/p>\n\n\n\n

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

        find the smallest number by which 132 must be multiplied so that the resulting number is a perfect cube The Correct Answer and Explanation is: To find the smallest number by which 132 must be multiplied so that the resulting number becomes a perfect cube, we need to analyze the prime factorization of 132. Step […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-25970","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/25970","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=25970"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/25970\/revisions"}],"predecessor-version":[{"id":25973,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/25970\/revisions\/25973"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=25970"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=25970"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=25970"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}