{"id":269011,"date":"2025-07-24T10:16:49","date_gmt":"2025-07-24T10:16:49","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=269011"},"modified":"2025-07-24T10:16:52","modified_gmt":"2025-07-24T10:16:52","slug":"show-that-the-product-of-3-consecutive-integers-is-divisible-by-6","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/24\/show-that-the-product-of-3-consecutive-integers-is-divisible-by-6\/","title":{"rendered":"Show that the product of 3 consecutive integers is divisible by 6."},"content":{"rendered":"\n

11 Let a,b, m \\in \\mathbb{Z}. Prove that (a,b) = (a, b+ma) 12 Show that the product of 3 consecutive integers is divisible by 6. 13. Let a, b \\in \\mathbb{Z}. Show that either (a+b) & (a-b) are even or both are odd.<\/p>\n\n\n\n

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

11. Prove that (a,b)=(a,b+ma)(a, b) = (a, b + ma)(a,b)=(a,b+ma) for a,b,m\u2208Za, b, m \\in \\mathbb{Z}a,b,m\u2208Z.<\/h3>\n\n\n\n

We are tasked with proving that the greatest common divisor (gcd) of aaa and bbb is the same as the gcd of aaa and b+mab + mab+ma, where m\u2208Zm \\in \\mathbb{Z}m\u2208Z.<\/p>\n\n\n\n

Proof:<\/strong><\/p>\n\n\n\n

Let d=(a,b)d = (a, b)d=(a,b), meaning ddd is the greatest common divisor of aaa and bbb. By the definition of gcd, this means:d\u2223aandd\u2223b.d \\mid a \\quad \\text{and} \\quad d \\mid b.d\u2223aandd\u2223b.<\/p>\n\n\n\n

We need to show that d=(a,b+ma)d = (a, b + ma)d=(a,b+ma).<\/p>\n\n\n\n

Since d\u2223ad \\mid ad\u2223a and d\u2223bd \\mid bd\u2223b, we know d\u2223mad \\mid mad\u2223ma because d\u2223ad \\mid ad\u2223a and m\u2208Zm \\in \\mathbb{Z}m\u2208Z. Therefore:d\u2223(b+ma)(since d\u2223b and d\u2223ma).d \\mid (b + ma) \\quad \\text{(since \\(d \\mid b\\) and \\(d \\mid ma\\))}.d\u2223(b+ma)(since d\u2223b and d\u2223ma).<\/p>\n\n\n\n

Thus, d\u2223(b+ma)d \\mid (b + ma)d\u2223(b+ma), meaning d\u2264(a,b+ma)d \\leq (a, b + ma)d\u2264(a,b+ma).<\/p>\n\n\n\n

Now, let e=(a,b+ma)e = (a, b + ma)e=(a,b+ma). This means that e\u2223ae \\mid ae\u2223a and e\u2223(b+ma)e \\mid (b + ma)e\u2223(b+ma). Since e\u2223ae \\mid ae\u2223a, we also have e\u2223mae \\mid mae\u2223ma (because e\u2223ae \\mid ae\u2223a and m\u2208Zm \\in \\mathbb{Z}m\u2208Z). Moreover, since e\u2223(b+ma)e \\mid (b + ma)e\u2223(b+ma), we can write:b+ma=k\u22c5efor some integer k.b + ma = k \\cdot e \\quad \\text{for some integer } k.b+ma=k\u22c5efor some integer k.<\/p>\n\n\n\n

Thus:b=k\u22c5e\u2212ma.b = k \\cdot e – ma.b=k\u22c5e\u2212ma.<\/p>\n\n\n\n

Since e\u2223ae \\mid ae\u2223a and e\u2223(k\u22c5e\u2212ma)e \\mid (k \\cdot e – ma)e\u2223(k\u22c5e\u2212ma), we have e\u2223be \\mid be\u2223b, which implies e\u2264de \\leq de\u2264d.<\/p>\n\n\n\n

Therefore, since d\u2264ed \\leq ed\u2264e and e\u2264de \\leq de\u2264d, we conclude that d=ed = ed=e. Hence:(a,b)=(a,b+ma),(a, b) = (a, b + ma),(a,b)=(a,b+ma),<\/p>\n\n\n\n

which completes the proof.<\/p>\n\n\n\n

\n\n\n\n

12. Show that the product of 3 consecutive integers is divisible by 6.<\/h3>\n\n\n\n

Let the three consecutive integers be n\u22121n-1n\u22121, nnn, and n+1n+1n+1, where n\u2208Zn \\in \\mathbb{Z}n\u2208Z.<\/p>\n\n\n\n

Proof:<\/strong><\/p>\n\n\n\n

The product of these integers is:(n\u22121)\u22c5n\u22c5(n+1).(n – 1) \\cdot n \\cdot (n + 1).(n\u22121)\u22c5n\u22c5(n+1).<\/p>\n\n\n\n

We need to prove that this product is divisible by 6. Notice that 6 can be factored as:6=2\u22c53.6 = 2 \\cdot 3.6=2\u22c53.<\/p>\n\n\n\n

For the product to be divisible by 6, it must be divisible by both 2 and 3.<\/p>\n\n\n\n

Divisibility by 2:<\/strong><\/p>\n\n\n\n

Among any three consecutive integers, at least one of them must be divisible by 2 (since every other integer is even). Therefore, the product is always divisible by 2.<\/p>\n\n\n\n

Divisibility by 3:<\/strong><\/p>\n\n\n\n

Among any three consecutive integers, at least one of them must be divisible by 3. This is because every third integer is divisible by 3. Therefore, the product is always divisible by 3.<\/p>\n\n\n\n

Since the product is divisible by both 2 and 3, it is divisible by 6. Hence, the product of any three consecutive integers is divisible by 6.<\/p>\n\n\n\n

\n\n\n\n

13. Let a,b\u2208Za, b \\in \\mathbb{Z}a,b\u2208Z. Show that either (a+b)(a + b)(a+b) and (a\u2212b)(a – b)(a\u2212b) are both even or both odd.<\/h3>\n\n\n\n

We are given a,b\u2208Za, b \\in \\mathbb{Z}a,b\u2208Z, and we need to prove that either both a+ba + ba+b and a\u2212ba – ba\u2212b are even, or both are odd.<\/p>\n\n\n\n

Proof:<\/strong><\/p>\n\n\n\n

Consider the parity (whether the number is odd or even) of aaa and bbb. There are two possibilities for each of aaa and bbb:<\/p>\n\n\n\n

    \n
  • aaa is even or odd.<\/li>\n\n\n\n
  • bbb is even or odd.<\/li>\n<\/ul>\n\n\n\n

    We will examine the four possible cases for aaa and bbb:<\/p>\n\n\n\n

      \n
    1. Case 1: aaa is even, bbb is even.<\/strong>
      • a+ba + ba+b is even (sum of two even numbers).<\/li>
      • a\u2212ba – ba\u2212b is even (difference of two even numbers).<\/li><\/ul>Hence, both a+ba + ba+b and a\u2212ba – ba\u2212b are even.<\/li>\n\n\n\n
      • Case 2: aaa is even, bbb is odd.<\/strong>
        • a+ba + ba+b is odd (sum of an even and an odd number).<\/li>
        • a\u2212ba – ba\u2212b is odd (difference of an even and an odd number).<\/li><\/ul>Hence, both a+ba + ba+b and a\u2212ba – ba\u2212b are odd.<\/li>\n\n\n\n
        • Case 3: aaa is odd, bbb is even.<\/strong>
          • a+ba + ba+b is odd (sum of an odd and an even number).<\/li>
          • a\u2212ba – ba\u2212b is odd (difference of an odd and an even number).<\/li><\/ul>Hence, both a+ba + ba+b and a\u2212ba – ba\u2212b are odd.<\/li>\n\n\n\n
          • Case 4: aaa is odd, bbb is odd.<\/strong>
            • a+ba + ba+b is even (sum of two odd numbers).<\/li>
            • a\u2212ba – ba\u2212b is even (difference of two odd numbers).<\/li><\/ul>Hence, both a+ba + ba+b and a\u2212ba – ba\u2212b are even.<\/li>\n<\/ol>\n\n\n\n

              In all four cases, we see that either both a+ba + ba+b and a\u2212ba – ba\u2212b are even, or both are odd. Therefore, the statement is proven.<\/p>\n\n\n\n

              \n\n\n\n

              This completes the proofs for all three questions.<\/p>\n\n\n\n

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

              11 Let a,b, m \\in \\mathbb{Z}. Prove that (a,b) = (a, b+ma) 12 Show that the product of 3 consecutive integers is divisible by 6. 13. Let a, b \\in \\mathbb{Z}. Show that either (a+b) & (a-b) are even or both are odd. The Correct Answer and Explanation is: 11. Prove that (a,b)=(a,b+ma)(a, b) = […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-269011","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/269011","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=269011"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/269011\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=269011"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=269011"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=269011"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}