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Re: Math: Number Theory
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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
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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
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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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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
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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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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
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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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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
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14 Feb 2015, 23:10
First of all, great post! Thanks a ton for creating this resource.
I had a very quick question.
In some DS questions, I have come accross the term  "Range of n integers"
I first assumed, range would mean the number of terms.
I was able to get some of the questions correct, using this but I think I got very lucky. Primarily because when I use different methods to check and practice the question, it leads me to a different answer.
Any chance you can let me know if my assumption was correct? If so, any suggestions how best to tackle these questions in the least amount of time.
Many thanks!



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Re: Math: Number Theory
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16 Feb 2015, 03:25
ud19 wrote: First of all, great post! Thanks a ton for creating this resource.
I had a very quick question.
In some DS questions, I have come accross the term  "Range of n integers"
I first assumed, range would mean the number of terms.
I was able to get some of the questions correct, using this but I think I got very lucky. Primarily because when I use different methods to check and practice the question, it leads me to a different answer.
Any chance you can let me know if my assumption was correct? If so, any suggestions how best to tackle these questions in the least amount of time.
Many thanks! The range of a set is the difference between the largest and the smallest numbers of a set. For example, the range of {1, 10, 12} is 12  1 = 11 and the range of {7, 0, 2, 9} is 9  (7) = 16. DS Statistics and Sets problems to practice: search.php?search_id=tag&tag_id=34PS Statistics and Sets problems to practice: search.php?search_id=tag&tag_id=55Hope it helps.
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Re: Math: Number Theory
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12 Jun 2015, 11:27
Bunuel wrote: • Verifying the primality (checking whether the number is a prime) of a given number \(n\) can be done by trial division, that is to say dividing \(n\) by all integer numbers smaller than \(\sqrt{n}\), thereby checking whether \(n\) is a multiple of \(m<\sqrt{n}\). Example: Verifying the primality of \(161\): \(\sqrt{161}\) is little less than \(13\), from integers from \(2\) to \(13\), \(161\) is divisible by \(7\), hence \(161\) is not prime. A minor point, but the inequalities here should not be strict. If you want to test if some large integer n is prime, then you need to try dividing by numbers up to and including \(\sqrt{n}\). We must include \(\sqrt{n}\), in case our number is equal to the square of a prime. And it might be worth mentioning that it is only necessary to try dividing by prime numbers up to \(\sqrt{n}\), since if n has any divisors at all (besides 1 and n), then it must have a prime divisor. It's very rare, though, that one needs to test if a number is prime on the GMAT. It is, computationally, extremely timeconsuming to test if a large number is prime, so the GMAT cannot ask you to do that. If a GMAT question asks if a large number is prime, the answer really must be 'no', because while you can often quickly prove a large number is not prime (for example, 1,000,011 is not prime because it is divisible by 3, as we see by summing digits), you cannot quickly prove that a large number is prime.
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12 Jun 2015, 11:52
IanStewart wrote: Bunuel wrote: • Verifying the primality (checking whether the number is a prime) of a given number \(n\) can be done by trial division, that is to say dividing \(n\) by all integer numbers smaller than \(\sqrt{n}\), thereby checking whether \(n\) is a multiple of \(m<\sqrt{n}\). Example: Verifying the primality of \(161\): \(\sqrt{161}\) is little less than \(13\), from integers from \(2\) to \(13\), \(161\) is divisible by \(7\), hence \(161\) is not prime. A minor point, but the inequalities here should not be strict. If you want to test if some large integer n is prime, then you need to try dividing by numbers up to and including \(\sqrt{n}\). We must include \(\sqrt{n}\), in case our number is equal to the square of a prime. And it might be worth mentioning that it is only necessary to try dividing by prime numbers up to \(\sqrt{n}\), since if n has any divisors at all (besides 1 and n), then it must have a prime divisor. It's very rare, though, that one needs to test if a number is prime on the GMAT. It is, computationally, extremely timeconsuming to test if a large number is prime, so the GMAT cannot ask you to do that. If a GMAT question asks if a large number is prime, the answer really must be 'no', because while you can often quickly prove a large number is not prime (for example, 1,000,011 is not prime because it is divisible by 3, as we see by summing digits), you cannot quickly prove that a large number is prime. Than you Ian. Edited the typo.
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Re: Math: Number Theory
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08 Aug 2015, 23:27
In Roots section you wrote:
\(\sqrt{x^2}\)=x, when x≤0, then, \(\sqrt{x^2}\)=x and when x≥0, then \(\sqrt{x^2}\)=x.
Here, won't it be "when x<0, then \(\sqrt{x^2}\)=−x"?



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16 Aug 2015, 10:11
zmtalha wrote: In Roots section you wrote:
\(\sqrt{x^2}\)=x, when x≤0, then, \(\sqrt{x^2}\)=x and when x≥0, then \(\sqrt{x^2}\)=x.
Here, won't it be "when x<0, then \(\sqrt{x^2}\)=−x"? No. For x = 0, we'd get \(\sqrt{0^2}=0=0\).
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18 Aug 2015, 13:58
how do you get the 13 and the 11 in the question? If the sum of two positive integers is 24 and the difference of their squares is 48, what is the product of the two integers?
x+y=24 and x2−y2=48 > (x+y)(x−y)=48, as x+y=24 > 24(x−y)=48 > x−y=2 > solving for x and y > x=13 and y=11 > xy=143.
Answer: E.



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23 Aug 2015, 21:40
I have a theoryrelated question regarding the definition of intersecting lines. If a line overlaps (colinear) a line segment (more than one point, of course) are they still considered to be "intersecting"? I was under the impression that this would not constitute an intersection however a question I worked seemed to suggest the opposite.
Example: line segment (1,5),(3,3) and line y=x+6
The line overlaps the line segment; are they "intersecting"?



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24 Aug 2015, 08:42
ar500 wrote: I have a theoryrelated question regarding the definition of intersecting lines. If a line overlaps (colinear) a line segment (more than one point, of course) are they still considered to be "intersecting"? I was under the impression that this would not constitute an intersection however a question I worked seemed to suggest the opposite.
Example: line segment (1,5),(3,3) and line y=x+6
The line overlaps the line segment; are they "intersecting"? Technically intersect means share one or more points in common. So, if two lines overlap they do intersect. Having said that, I must add that GMAT would never test you on such technicalities, you can ignore this question and move on. Hope it helps.
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14 Sep 2016, 06:53
Quote: Similarly, all prime numbers above 3 are of the form 6n−16n−1 or 6n+16n+1, because all other numbers are divisible by 2 or 3. In the prime number section it says prime numbers greater than 3 are of the form 6n1 or 6n+1. However, this is not necessarily true. eg: n=36, then 6n1 is 215 and 6n+1 is 217, divisible by 5 and 7 respectively.



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14 Sep 2016, 07:02
ks4196 wrote: Quote: Similarly, all prime numbers above 3 are of the form 6n−16n−1 or 6n+16n+1, because all other numbers are divisible by 2 or 3. In the prime number section it says prime numbers greater than 3 are of the form 6n1 or 6n+1. However, this is not necessarily true. eg: n=36, then 6n1 is 215 and 6n+1 is 217, divisible by 5 and 7 respectively. That's not what is said there. First of all there is no known formula of prime numbers.Next:Any prime number \(p>3\) when divided by 6 can only give remainder of 1 or 5 (remainder can not be 2 or 4 as in this case \(p\) would be even and remainder can not be 3 as in this case \(p\) would be divisible by 3). So any prime number \(p>3\) could be expressed as \(p=6n+1\) or\(p=6n+5\) or \(p=6n1\), where n is an integer >1. But:Not all number which yield a remainder of 1 or 5 upon division by 6 are prime, so viseversa of above property is not correct. For example 25 yields a remainder of 1 upon division be 6 and it's not a prime number. Hope it's clear.
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Re: Math: Number Theory &nbs
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14 Sep 2016, 07:02



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