{"id":5431,"date":"2015-04-30T08:12:11","date_gmt":"2015-04-30T15:12:11","guid":{"rendered":"https:\/\/e-gmat.com\/blogs\/?p=5431"},"modified":"2015-04-30T21:00:39","modified_gmt":"2015-05-01T04:00:39","slug":"3-mistakes-of-lcm-gcd-questions-q6","status":"publish","type":"post","link":"https:\/\/e-gmat.com\/blogs\/3-mistakes-of-lcm-gcd-questions-q6\/","title":{"rendered":"3 mistakes of LCM-GCD questions Q6"},"content":{"rendered":"A <\/span> 2<\/span> min read <\/span><\/span>

Question 6<\/h2>\n

Q6: For two positive integers A & B, what is the highest number that divides completely the product of integers from 1 to A and 1 to B such that B = A + 29<\/em><\/strong>.<\/p>\n

    \n
  1. 1<\/strong><\/li>\n
  2. Product of all integers from 1 to A<\/strong><\/li>\n
  3. Product of all integers from 1 to B<\/strong><\/li>\n
  4. 29*A<\/strong><\/li>\n
  5. Can’t be determined<\/strong><\/li>\n<\/ol>\n

    Correct Answer<\/h3>\n

    B<\/p>\n

    Solution<\/h3>\n

    Given<\/strong><\/span><\/p>\n

    We are given two positive integers A & B. We are asked to find the highest number which divides product of all integers from 1 to A product of all integers from 1 to B. We are also given that B = A +29<\/p>\n

     <\/p>\n

    Approach<\/strong><\/span><\/p>\n

    The highest number which divides two numbers is the GCD of the numbers. So the question is asking us to find the GCD of two numbers.<\/p>\n

     <\/p>\n

    The first number is the product of all integers from 1 to A. This product is known as A Factorial and is represented A!<\/p>\n

    We can write A! = 1* 2* 3\u2026\u2026A.<\/p>\n

    Similarly product of all integers from 1 to B is known as B!<\/p>\n

    We can write B! = 1* 2* 3\u2026\u2026\u2026\u2026..B.<\/p>\n

     <\/p>\n

    So, we are asked to find the GCD (A! , B!). Since, we are given that B = A + 29 it would imply that B > A<\/p>\n

    Let\u2019s use our understanding of GCD to find out the GCD (A! , B!)<\/p>\n

     <\/p>\n

    Working Out<\/strong><\/span><\/p>\n

    Since B > A, we can write B! = 1* 2* 3\u2026..A<\/span> * (A+1)*(A+2)\u2026\u2026B.<\/p>\n

    We know that 1*2*3\u2026.. A = A!.<\/p>\n

     <\/p>\n

    So, B! = A!<\/span> * (A+1)*(A+2)\u2026B.<\/p>\n

    We see here that A! is a factor of both A! and B!. We also know that GCD of a set of numbers can\u2019t be greater than the magnitude of the smallest number which in this case is A!<\/p>\n

     <\/p>\n

    Hence, we can say that A! is the highest number which divides both A! and B!. Therefore GCD(A! , B!) = A! i.e. product of all integers from 1 to A<\/p>\n

     <\/p>\n

    Answer: Option (B)<\/strong><\/p>\n


    \n<\/strong><\/p>\n

     <\/p>\n


    \n<\/strong><\/p>\n

     <\/p>\n


    \n<\/strong><\/p>\n

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

    Question 6 Q6: For two positive integers A & B, what is the highest number that divides completely the product of integers from 1 to A and 1 to B such that B = A + 29. 1 Product of all integers from 1 to A Product of all integers from 1 to B 29*A […]<\/p>\n","protected":false},"author":7,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_et_pb_use_builder":"","_et_pb_old_content":"","_et_gb_content_width":"","ub_ctt_via":""},"categories":[1],"tags":[],"featured_image_src":null,"author_info":{"display_name":"Krishna Chaitanya","author_link":"https:\/\/e-gmat.com\/blogs\/author\/krishnae-gmat-com\/"},"yoast_head":"\n3 mistakes of LCM-GCD questions Q6<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/e-gmat.com\/blogs\/3-mistakes-of-lcm-gcd-questions-q6\/\" \/>\n<meta property=\"og:locale\" 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