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| 1. |
Show that pp will a remainder when divided by 8 if p isan odd positive integer |
| Answer» Since p is an odd positive integer.Therefore, p = 2q\xa0+ 1 or 2q + 3 for some integer q.case 1: p = 2q + 1squaring both sides, we get,p2= (2q+1){tex}^2{/tex}=4q2\xa0+ 4q + 1= 4\xa0q(q+ 1)\xa0+ 1 Since we know that q(q+1) is even\xa0therefore , q(q+1)=2m{tex}\\implies {/tex}p{tex}^2 {/tex}\xa0={tex}4\\times2m+1{/tex}\xa0p{tex}^2 {/tex}\xa0=8m+1{tex}\\Rightarrow{/tex}\xa0p2\xa0leaves remainder 1 when divided by 8.case 2: p = 2q +\xa03p{tex}^2{/tex}=(2q+3){tex}^2{/tex}{tex}p^2\xa0= 4q^2\xa0+ 12q\xa0+ 9{/tex}{tex}p^2\xa0= 4q^2\xa0+ 12q\xa0+ 8 + 1{/tex}= 4q(q+3)+8+1\xa0={tex}4\\times2n+8+1{/tex}(since q(q+3) is even number, so q(q+3)=2n\xa0= 8n+8+1\xa0=8(n+1)+1 =8m+1, where m=n+1{tex}{/tex}{tex}\\therefore{/tex}p2\xa0leaves a remainder 1 when divided by 8. | |