Hello Sir, thanks a lot for ur essential help in prepping. Just one query, can we extend the theory for NONperfect 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:If N is a perfect square, then the number of factors of N will ALWAYS be an ODD number.
If N is a NONperfect square, then the number of factors of N will ALWAYS be an EVEN number.
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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) / a1 ] * [ (b^(q+1)  1) / b1 ] * [ (c^(r+1)  1) / c1 ]
Thanks a ton for this utterly useful formula. This is just awesome.

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

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But the fun part is proving it!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) / a1 ] * [ (b^(q+1)  1) / b1 ] * [ (c^(r+1)  1) / c1 ]
Thanks a ton for this utterly useful formula. This is just awesome.
How would we know that such a formula works?

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Piggybacking on my last post, a more memorable formula (in my opinion) is
(aáµ– + aáµ–â�»Â¹ + ... + aÂ¹ + 1) * (báµ’ + báµ’â�»Â¹ + ... + bÂ¹ + 1) * (cÊ³ + cÊ³â�»Â¹ + ... + cÂ¹ + 1)
I prefer this one  it shows a little more clearly where the formula comes from.
(aáµ– + aáµ–â�»Â¹ + ... + aÂ¹ + 1) * (báµ’ + báµ’â�»Â¹ + ... + bÂ¹ + 1) * (cÊ³ + cÊ³â�»Â¹ + ... + cÂ¹ + 1)
I prefer this one  it shows a little more clearly where the formula comes from.

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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.

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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.

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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.

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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.

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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 3Â²Â³. We know that the units digits cycle in a block of 4: 3, 9, 7, 1, 3, 9, 7, 1, ...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.
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 3Â²Â³ ends in 7. At the risk of sounding like the dorkiest math teacher you had ... too cool!

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