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Re: Math: Number Theory [#permalink]
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27 Oct 2012, 02:00
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I shall like to add one trick to find square of number : To find square of a number \((ab)^2\) where a is tens digit and b is units digit => 1. Multiply tens digit a by (a+1) i.e. a * (a+1) = A 2. Suffix 25 to the above derived result A and the new number will be A25 Example: what is \(35^2\)? Solution: 3 * (3+1) = 3*4 = 12 and so answer is 1225
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Re: Math: Number Theory [#permalink]
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27 Oct 2012, 02:34
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About Exponents and divisibility: \((a + b)^2 = a^2+ 2ab + b^2\) Square of a Sum \((a  b)^2 = a^2  2ab + b^2\) Square of a Difference \(a^n  b^n\) is always divisble by ab i.e. irrespective of n being odd or even Proof: \(a^2  b^2 = (ab)(a+b)\) \(a^3  b^3 = (ab)(a^2+ab+b^2)\) Thus divisible by a b in both cases where n = 2 i.e. even and 3 i.e. odd \(a^n + b^n\) is divisble by a+b i.e. only if n = odd Proof: \(a^3  b^3 = (a+b)(a^2ab+b^2)\) Thus divisible by a + b as n = 3 i.e. odd
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Re: Math: Number Theory [#permalink]
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12 Dec 2012, 03:26
Hi,
I'm not sure whether I undertood the below rule correctly:
"Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base".
55^2 = 3025  the last digit is same as the base (5) so the above rule works. 55^10 = 253295162119141000  the last digit is not same as the base (5) so the above rule doesn't work.
Please help if I have misunderstood the rule.



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Re: Math: Number Theory [#permalink]
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12 Dec 2012, 03:33
klueless7825 wrote: Hi,
I'm not sure whether I undertood the below rule correctly:
"Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base".
55^2 = 3025  the last digit is same as the base (5) so the above rule works. 55^10 = 25329516211914100[b]0[/b]  the last digit is not same as the base (5) so the above rule doesn't work.
Please help if I have misunderstood the rule. 5 in any positive integer power has 5 as the units digit. 5^1=5; 5^2=25; 5^3=125 ... 5^10=253,295,162,119,140,625 (your result was just rounded). Hope it's clear.
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Re: Math: Number Theory [#permalink]
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25 Dec 2012, 08:33
Thanks for the post!



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Re: Math: Number Theory [#permalink]
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12 Jul 2013, 03:12
Great Post. thanks a lot



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Re: Math: Number Theory [#permalink]
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15 Sep 2013, 10:04
Hi!!! can you explain the number line?? and how we can figure out the radicals on it???



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02 Oct 2013, 20:39
Awesome post. Bunuel, you are great. Love your posts.
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Re: Math: Number Theory [#permalink]
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14 Oct 2013, 10:47
can someone explain me this property:
If a is a factor of bc, and gcd(a,b)=1, then a is a factor of c.
???



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Re: Math: Number Theory [#permalink]
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14 Oct 2013, 11:18
Quite informative and descriptive... all at one place



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Re: Math: Number Theory [#permalink]
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17 Oct 2013, 03:33
skamran wrote: can someone explain me this property:
If a is a factor of bc, and gcd(a,b)=1, then a is a factor of c.
??? Say a=2, b=3 (gcd(a,b)=gcd(2,3)=1), and c=4. a=2 IS a factor of bc=12, and a=2 IS a factor of c. OR: if a is a factor of bc and NOT a factor of b, then it must be a factor of c. Hope it's clear.
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Re: Math: Number Theory [#permalink]
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17 Oct 2013, 16:52
Bunuel wrote: skamran wrote: can someone explain me this property:
If a is a factor of bc, and gcd(a,b)=1, then a is a factor of c.
??? Say a=2, b=3 (gcd(a,b)=gcd(2,3)=1), and c=4. a=2 IS a factor of bc=12, and a=2 IS a factor of c. OR: if a is a factor of bc and NOT a factor of b, then it must be a factor of c. Hope it's clear. Yeh Thanks alot!!!



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Re: Math: Number Theory [#permalink]
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29 Oct 2013, 21:06
Hello Bunuel,
• \sqrt{x^2}=x, when x\leq{0}, then \sqrt{x^2}=x and when x\geq{0}, then \sqrt{x^2}=x
• When the GMAT provides the square root sign for an even root, such as \sqrt{x} or \sqrt[4]{x}, then the only accepted answer is the positive root.
Isn't both of these points contradict each other?
If I consider second point as valid then how can \sqrt{x^2}=x, when x\leq{0}, then \sqrt{x^2}=x be said ?



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Re: Math: Number Theory [#permalink]
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30 Oct 2013, 00:31
Phoenix22 wrote: Hello Bunuel,
• \sqrt{x^2}=x, when x\leq{0}, then \sqrt{x^2}=x and when x\geq{0}, then \sqrt{x^2}=x
• When the GMAT provides the square root sign for an even root, such as \sqrt{x} or \sqrt[4]{x}, then the only accepted answer is the positive root.
Isn't both of these points contradict each other?
If I consider second point as valid then how can \sqrt{x^2}=x, when x\leq{0}, then \sqrt{x^2}=x be said ? The point here is that square root function can not give negative result: wich means that \(\sqrt{some \ expression}\geq{0}\). So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to? Let's consider following examples: If \(x=5\) > \(\sqrt{x^2}=\sqrt{25}=5=x=positive\); If \(x=5\) > \(\sqrt{x^2}=\sqrt{25}=5=x=positive\). So we got that: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=x\), if \(x<0\). What function does exactly the same thing? The absolute value function! That is why \(\sqrt{x^2}=x\)
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Re: Math: Number Theory [#permalink]
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06 Nov 2013, 18:51
this is insanely helpful..thank you so much!!



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Re: Math: Number Theory [#permalink]
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13 Jan 2014, 20:57
Bunuel wrote: LAST DIGIT OF A POWER
Determining the last digit of \((xyz)^n\):
1. Last digit of \((xyz)^n\) is the same as that of \(z^n\); 2. Determine the cyclicity number \(c\) of \(z\); 3. Find the remainder \(r\) when \(n\) divided by the cyclisity; 4. When \(r>0\), then last digit of \((xyz)^n\) is the same as that of \(z^r\) and when \(r=0\), then last digit of \((xyz)^n\) is the same as that of \(z^c\), where \(c\) is the cyclisity number.
• Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base. • Integers ending with 2, 3, 7 and 8 have a cyclicity of 4. • Integers ending with 4 (eg. \((xy4)^n\)) have a cyclisity of 2. When n is odd \((xy4)^n\) will end with 4 and when n is even \((xy4)^n\) will end with 6. • Integers ending with 9 (eg. \((xy9)^n\)) have a cyclisity of 2. When n is odd \((xy9)^n\) will end with 9 and when n is even \((xy9)^n\) will end with 1.
Example: What is the last digit of \(127^{39}\)? Solution: Last digit of \(127^{39}\) is the same as that of \(7^{39}\). Now we should determine the cyclisity of \(7\):
1. 7^1=7 (last digit is 7) 2. 7^2=9 (last digit is 9) 3. 7^3=3 (last digit is 3) 4. 7^4=1 (last digit is 1) 5. 7^5=7 (last digit is 7 again!) ...
So, the cyclisity of 7 is 4.
Now divide 39 (power) by 4 (cyclisity), remainder is 3.So, the last digit of \(127^{39}\) is the same as that of the last digit of \(7^{39}\), is the same as that of the last digit of \(7^3\), which is \(3\).
Congratulation and thank you very much for the post, but in the LAST DIGIT OF A POWER i have an issue, when i try to solve the last digit of (456)^35 with the process i just don't get the correct answers, with the process above gives me 6^4 which is 1296=6 and with calculator its 0, can you explain me that case?



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Re: Math: Number Theory [#permalink]
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14 Jan 2014, 01:56
mandrake15 wrote: Bunuel wrote: LAST DIGIT OF A POWER
Determining the last digit of \((xyz)^n\):
1. Last digit of \((xyz)^n\) is the same as that of \(z^n\); 2. Determine the cyclicity number \(c\) of \(z\); 3. Find the remainder \(r\) when \(n\) divided by the cyclisity; 4. When \(r>0\), then last digit of \((xyz)^n\) is the same as that of \(z^r\) and when \(r=0\), then last digit of \((xyz)^n\) is the same as that of \(z^c\), where \(c\) is the cyclisity number.
• Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base. • Integers ending with 2, 3, 7 and 8 have a cyclicity of 4. • Integers ending with 4 (eg. \((xy4)^n\)) have a cyclisity of 2. When n is odd \((xy4)^n\) will end with 4 and when n is even \((xy4)^n\) will end with 6. • Integers ending with 9 (eg. \((xy9)^n\)) have a cyclisity of 2. When n is odd \((xy9)^n\) will end with 9 and when n is even \((xy9)^n\) will end with 1.
Example: What is the last digit of \(127^{39}\)? Solution: Last digit of \(127^{39}\) is the same as that of \(7^{39}\). Now we should determine the cyclisity of \(7\):
1. 7^1=7 (last digit is 7) 2. 7^2=9 (last digit is 9) 3. 7^3=3 (last digit is 3) 4. 7^4=1 (last digit is 1) 5. 7^5=7 (last digit is 7 again!) ...
So, the cyclisity of 7 is 4.
Now divide 39 (power) by 4 (cyclisity), remainder is 3.So, the last digit of \(127^{39}\) is the same as that of the last digit of \(7^{39}\), is the same as that of the last digit of \(7^3\), which is \(3\).
Congratulation and thank you very much for the post, but in the LAST DIGIT OF A POWER i have an issue, when i try to solve the last digit of (456)^35 with the process i just don't get the correct answers, with the process above gives me 6^4 which is 1296=6 and with calculator its 0, can you explain me that case? Any integer with 6 as its units digit in any positive integer power has the units digit of 6 (integers ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base.). For example, (xxx6)^(positive integer) has the units digit of 6. The reason you get 0 as the units digit of (456)^35 is because it's a huge number and simple calculator rounds the result. Exact result is: 1,158,162,485,059,181,044,784,824,077,056,791,483,879,723,809,565,243,305,114,019,731,744,476,935,058,125,438,332,149,170,176. 1 trigintillion 158 novemvigintillion 162 octovigintillion 485 septenvigintillion 59 sexvigintillion 181 quinvigintillion 44 quattuorvigintillion 784 trevigintillion 824 duovigintillion 77 unvigintillion 56 vigintillion 791 novemdecillion 483 octodecillion 879 septendecillion 723 sexdecillion 809 quindecillion 565 quattuordecillion 243 tredecillion 305 duodecillion 114 undecillion 19 decillion 731 nonillion 744 octillion 476 septillion 935 sextillion 58 quintillion 125 quadrillion 438 trillion 332 billion 149 million 170 thousand 176 Hope it's clear.
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Re: Math: Number Theory [#permalink]
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09 Feb 2014, 09:07
any body can explain that why we need to find the sum of number's factors ????




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