# find the smallest number which leave remember 8 and 12 when divided by 28 and 32 respectively

$We,\;know\;that\;Divident=Divisor\times Quotient+Remainder\\* So,\;when\;we\;divide\;a\;number\;by\;28\;and\;32\;we\;can\;write\;it\;as\\* 28\times p+8\; and\;32\times q+12\; and\; they\;both\;should\;be\;equal.\\* So,\\* \Rightarrow 28\times p+8=32\times q+12\\*\indent where, p,q\;are\;natural\;numbers.\;\\* \Rightarrow 28\times p+8-12=32\times q\\* \Rightarrow q=\frac{28p-4}{32}=\frac{7p-1}{8}.....eq.(1)\\* Now, \;by\;trial\;and\;error\;method\\* we\;see\;that\;q=7,\;will\;satisfy\;the\;eq.(1)\\* Now,\;put\;q=7, \;and \;we\;get\;r=6\\* both\;are\;natural\;numbers.\\* So,\\*\indent the\;smallest\;number=28\times 7 +8=204\\* Therefore, \; smallest\;number\;which\;leaves\;remainder\;8\;and\;12\;when\\*divided\;by\;28\;and\;32\;respectively\;is\;204.$

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