papgust wrote:If N is a perfect square, then the number of factors of N will ALWAYS be an ODD number.

If N is a NON-perfect square, then the number of factors of N will ALWAYS be an EVEN number.

Hello Sir, thanks a lot for ur essential help in prepping. Just one query, can we extend the theory for NON-perfect square to "all positive integers except squares"? Cos it's just one odd power that's needed to make the number of factors even. Pls give ur insight.

papgust wrote:How to find Sum of all factors of a POSITIVE integer:

If N is expressed in terms of its prime factors as a^p * b^q * c^r, where p,q,r are positive integers, then the sum of all factors of N is

[ (a^(p+1) - 1) / a-1 ] * [ (b^(q+1) - 1) / b-1 ] * [ (c^(r+1) - 1) / c-1 ]

Thanks a ton for this utterly useful formula. This is just awesome.

Very nice post and right to the point. I am not sure if this is actually the best place to ask but do you folks have any idea where to hire some professional writers? Thanks in advance

run3 wrote:Very nice post and right to the point. I am not sure if this is actually the best place to ask but do you folks have any idea where to hire some professional writers? Thanks in advance

What sort of writer? There are so many freelancers out there, and plenty of sites on which they advertise.

binit wrote:

papgust wrote:How to find Sum of all factors of a POSITIVE integer:

If N is expressed in terms of its prime factors as a^p * b^q * c^r, where p,q,r are positive integers, then the sum of all factors of N is

[ (a^(p+1) - 1) / a-1 ] * [ (b^(q+1) - 1) / b-1 ] * [ (c^(r+1) - 1) / c-1 ]

Thanks a ton for this utterly useful formula. This is just awesome.

But the fun part is proving it!

How would we know that such a formula works?

Piggybacking on my last post, a more memorable formula (in my opinion) is

(aᵖ + aᵖ�¹ + ... + a¹ + 1) * (bᵒ + bᵒ�¹ + ... + b¹ + 1) * (cʳ + cʳ�¹ + ... + c¹ + 1)

I prefer this one - it shows a little more clearly where the formula comes from.

drkomal2000 wrote:

papgust wrote: Pattern 1:
Unit's place that has digits - 2/3/7/8

Then, unit's digit repeats every 4th value. Divide the power (or index) by 4.

Hi Papgust
Thanks a lot for your notes. Can you or anyone from the community give me example of the pattern 1? I know you gave one example but Its confusing for me.

powers of 2: 2, 4, 8, 16, 32, 64, 128, 256, ...

drkomal2000 wrote:

papgust wrote: Pattern 1:
Unit's place that has digits - 2/3/7/8

Then, unit's digit repeats every 4th value. Divide the power (or index) by 4.

Hi Papgust
Thanks a lot for your notes. Can you or anyone from the community give me example of the pattern 1? I know you gave one example but Its confusing for me.

powers of 3: 3, 9, 27, 81, 243, 729, 2187, 6561, ...

drkomal2000 wrote:

papgust wrote: Pattern 1:
Unit's place that has digits - 2/3/7/8

Then, unit's digit repeats every 4th value. Divide the power (or index) by 4.

Hi Papgust
Thanks a lot for your notes. Can you or anyone from the community give me example of the pattern 1? I know you gave one example but Its confusing for me.

powers of 7: 7, 49, 343, 2401, 16807, 117649, 823543, 5764801, ...

drkomal2000 wrote:

papgust wrote: Pattern 1:
Unit's place that has digits - 2/3/7/8

Then, unit's digit repeats every 4th value. Divide the power (or index) by 4.

Hi Papgust
Thanks a lot for your notes. Can you or anyone from the community give me example of the pattern 1? I know you gave one example but Its confusing for me.

powers of 8: 8, 64, 512, 4096, 32768, 262144, 2097152, 16777216, ...

drkomal2000 wrote:

papgust wrote: Pattern 1:
Unit's place that has digits - 2/3/7/8

Then, unit's digit repeats every 4th value. Divide the power (or index) by 4.

Hi Papgust
Thanks a lot for your notes. Can you or anyone from the community give me example of the pattern 1? I know you gave one example but Its confusing for me.

Once we're armed with those, we can use the powers of an integer to determine the units digit. For instance, suppose I ask for the units digit of 3²³. We know that the units digits cycle in a block of 4: 3, 9, 7, 1, 3, 9, 7, 1, ...

With that in mind, we only need to know the remainder when 23 is divided by 4. That remainder is 3, meaning that its units digit will be the third in the cycle {3, 9, 7, 1} and that 3²³ ends in 7. At the risk of sounding like the dorkiest math teacher you had ... too cool!

Thank you so much for the flashcards. They are superb and awesome...