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| 1. |
If p is the HCF of 45 and 27 , find a and b such that p= 27a+45b |
| Answer» 45 = 27 {tex} \\times {/tex}\xa01 + 1827 =18 {tex} \\times {/tex}\xa01 + 918 = 9 {tex} \\times {/tex}\xa02 + 0So H.C.F. =9Now 9 = 27 – 18 {tex} \\times {/tex}\xa01{tex}\\style{font-family:Arial}{\\begin{array}{l}9=27-18\\\\=27-(45-27\\times1)\\times1\\\\=27-45+27\\\\=2\\times27-(1)\\times45=27x+45y\\end{array}}{/tex}⇒ a\xa0= 2, b\xa0= – 1. | |