Math: Number Theory

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.

???

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 ?

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?

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!

• 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 time-consuming 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.

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"?

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.

I have a theory-related 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"?

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 6n-1 or 6n+1. However, this is not necessarily true. eg: n=36, then 6n-1 is 215 and 6n+1 is 217, divisible by 5 and 7 respectively.