Solution 3.1:7c - Revision history

Tek at 14:22, 30 September 2008

Tek — Tue, 30 Sep 2008 14:22:47 GMT

←Older revision		Revision as of 14:22, 30 September 2008
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-	If we divide up	+	If we divide up 153 and 68 into their smallest possible integer factors, we can see whether there are squared numbers which can be taken out from under the roots, and if the expression can be simplified further,
-	~~<math>\text{~~153 ~~}</math>~~	+
-	and	+
-	~~<math>\text{~~68 ~~}</math>~~	+
-	into their smallest possible integer factors, we can see whether there are squared numbers which can be taken out from under the roots, and if the expression can be simplified further,	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& \text{153 }=3\centerdot 51=3\centerdot 3\centerdot 17=3^{2}\centerdot 17, \\~~	+
-	~~& 68=2\centerdot 37=2\centerdot 2\centerdot 17=2^{2}\centerdot 17 \\~~	+
-	~~\end{align}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	153 &= 3\cdot 51 = 3\cdot 3\cdot 17 = 3^{2}\cdot 17,\\[5pt]
		+	68 &= 2\cdot 37 = 2\cdot 2\cdot 17 = 2^{2}\cdot 17\,\textrm{.}
		+	\end{align}</math>}}

	We get		We get

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	\sqrt{153}-\sqrt{68}
-	& \sqrt~~{\text~~{153}~~}\text{~~-}\sqrt{68}~~\text{ }~~=\sqrt{3^{2}\~~centerdot~~ 17}-\sqrt{2^{2}\~~centerdot~~ 17} \\	+	&= \sqrt{3^{2}\cdot 17}-\sqrt{2^{2}\cdot 17}\\[5pt]
-	& =3\sqrt{17}-2\sqrt{17}=\sqrt{17} \\	+	&= 3\sqrt{17}-2\sqrt{17}\\[5pt]
-	\end{align}</math>	+	&= \sqrt{17}\,\textrm{.}
		+	\end{align}</math>}}


-	~~NOTE~~: A good tip for when we investigate whether an integer is divisible by	+	Note: A good tip for when we investigate whether an integer is divisible by 3 is to look at the sum of the number's digits. A number is divisible by 3 if and only if the sum of its digits is divisible by 3. E.g. the number 97818 is divisible by 3 because the sum of its digits, <math>9+7+8+1+8=33</math>, is divisible by 3 and, conversely, the number 11536 is not divisible by 3 since sum of its digits,
-	~~<math>\text{~~3~~}</math>~~	+	<math>1+1+5+3+6=16</math>, is not divisible by 3.
-	is to look at the sum of the number's digits. A number is divisible by	+
-	~~<math>\text{~~3~~}</math>~~	+
-	if and only if the sum of its digits is divisible by	+
-	~~<math>\text{~~3~~}</math>~~. E.g. the number	+
-	~~<math>\text{~~97818~~}</math>~~	+
-	is divisible by	+
-	~~<math>\text{~~3~~}</math>~~	+
-	because the sum of its digits,	+
-	<math>~~\text{~~9}+~~\text{~~7}+~~\text{~~8}+~~\text{~~1}+~~\text{~~8}=~~\text{~~33}</math>, is divisible by	+
-	~~<math>\text{~~3~~}</math>~~	+
-	and, conversely, the number	+
-	~~<math>\text{~~11536 ~~}</math>~~	+
-	is not divisible by	+
-	~~<math>\text{~~3~~}</math>~~	+
-	since sum of its digits,	+
-	<math>~~\text{~~1}+~~\text{~~1}+~~\text{~~5}+~~\text{~~3}+~~\text{~~6}=~~\text{~~16}</math> , is not divisible by	+
-	~~<math>\text{~~3~~}</math>~~.	+

Ian at 10:23, 23 September 2008

Ian — Tue, 23 Sep 2008 10:23:26 GMT

←Older revision		Revision as of 10:23, 23 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	If we divide up
-	<~~center~~> ~~[[Image~~:~~3_1_7c~~.~~gif]]~~ </~~center~~>	+	<math>\text{153 }</math>
-	{{~~NAVCONTENT_STOP~~}}	+	and
		+	<math>\text{68 }</math>
		+	into their smallest possible integer factors, we can see whether there are squared numbers which can be taken out from under the roots, and if the expression can be simplified further,
		+
		+
		+	<math>\begin{align}
		+	& \text{153 }=3\centerdot 51=3\centerdot 3\centerdot 17=3^{2}\centerdot 17, \\
		+	& 68=2\centerdot 37=2\centerdot 2\centerdot 17=2^{2}\centerdot 17 \\
		+	\end{align}</math>
		+
		+
		+	We get
		+
		+
		+	<math>\begin{align}
		+	& \sqrt{\text{153}}\text{-}\sqrt{68}\text{ }=\sqrt{3^{2}\centerdot 17}-\sqrt{2^{2}\centerdot 17} \\
		+	& =3\sqrt{17}-2\sqrt{17}=\sqrt{17} \\
		+	\end{align}</math>
		+
		+
		+	NOTE: A good tip for when we investigate whether an integer is divisible by
		+	<math>\text{3}</math>
		+	is to look at the sum of the number's digits. A number is divisible by
		+	<math>\text{3}</math>
		+	if and only if the sum of its digits is divisible by
		+	<math>\text{3}</math>. E.g. the number
		+	<math>\text{97818}</math>
		+	is divisible by
		+	<math>\text{3}</math>
		+	because the sum of its digits,
		+	<math>\text{9}+\text{7}+\text{8}+\text{1}+\text{8}=\text{33}</math>, is divisible by
		+	<math>\text{3}</math>
		+	and, conversely, the number
		+	<math>\text{11536 }</math>
		+	is not divisible by
		+	<math>\text{3}</math>
		+	since sum of its digits,
		+	<math>\text{1}+\text{1}+\text{5}+\text{3}+\text{6}=\text{16}</math> , is not divisible by
		+	<math>\text{3}</math>.

Tekbot: Lösning 3.1:7c moved to Solution 3.1:7c: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:15:43 GMT

Lösning 3.1:7c moved to Solution 3.1:7c: Robot: moved page

←Older revision

Revision as of 09:15, 9 September 2008

Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

Tekbot — Thu, 21 Aug 2008 06:54:59 GMT

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Oskar — Tue, 01 Apr 2008 14:13:54 GMT

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