NeoPets Riddles (Lenny Conundrums) and Answers Here

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seems we all agree i also got 8 and two ways

Unfortunately, no.

If it were a product instead of a sum, you would be all set. However, (a + b)^n = a^n + n * a^(n-1) * b + n * (n-1) / 2 * a^(n-2) * b^2 + ... + n! / ((n-m)! * m!) * a^(n - m) * b^m + ... + b^n

I got the result from [9(7)+2](3) if you do all that separate got:
9(7)=4782969 and thats=9
(9+2)3=11(3)=1331=8
and then you got 8

One of the uses of the described "digital sum" is to determine whether a number is divisible by 9. If the sum is divisible by 9 (or by 3), then the original number is divisible by 9 (or by 3). For instance:

. . . . .18: 1 + 8 = 9, which is divisible by 9 (and by 3)

. . . . .243: 2 + 4 + 3 = 9, which is divisible by 9 (and by 3)

...and so forth.

Obviously, any power of 9 is divisible by 9, so 9^x is always going to have a digital sum of 9. (In this case, x = 62773.)

Now consider any number of the form 9^x + 2, when raised to some odd power, 2n + 1. ("Odd" because 83721 is an odd number.) The expanded form, by the Binomial Theorem, will be as follows:

. . .(9^x + 2)^(2n + 1)

. . . . .= (9^x)^(2n + 1) + (binomial coefficient)((9^x)^(2n))(2^1) + ...

. . . . . . .+ (binomial coefficient)((9^x)^1)(2^(2n)) + 2^(2n + 1)

. . . . .= 9[(9^x)^(2n) + ... + (bin. coeff.)(9^1)(2^(2n))] + 2^(2n + 1)

. . . . .= 9[(stuff)] + (2^(2n))(2^1)

. . . . .= 9[(stuff)] + 2((2^2)^n)

. . . . .= 9[(stuff)] + 2(4^n)

Since anything with a factor of 9 is, by definition, divisible by 9, then the digital sum of the "9[(stuff)]" part is going to be "9". The "2(4^n)" part is twice some power of four. What will this give us?

. . . . .2(4^1) = 8

. . . . .2(4^2) = 2(16) = 32; 3 + 2 = 5

. . . . .2(4^3) = 2(64) = 128; 1 + 2 + 8 = 11, 1 + 1 = 2

. . . . .2(4^4) = 2(256) = 512; 5 + 1 + 2 = 8

. . . . .2(4^5) = 2(1024) = 2048; 2 + 0 + 4 + 8 = 14, 1 + 4 = 5

. . . . .2(4^6) = 2(4096) = 8192; 8 + 1 + 9 + 2 = 20, 2 + 0 = 2

So the digital sums appear to cycle as 8, 5, 2, 8, 5, 2,.... Thus, we need to determine where, in the cycle, 83721 appears. Since 83721 is odd, then 83721 = 2n + 1 for some value of n. Solving, we get:

. . . . .83721 = 2n + 1

. . . . .83720 = 2n

. . . . .41860 = n

Since the cycle of powers repeats every three powers (1, 2, 3, then 4, 5, 7, etc), we need to determine where 41860 falls.

. . . . .41860 = 41859 + 1 = 3(13953) + 1

That is, 41860 is one more than a multiple of 3. This puts the power of 83721 at the "8" in the cycle of digital sums for the 2(4^n) term.

Then we should get a sum, from the "9[(stuff)]" part, of "9", and a digital sum for the remaining term of "8". But are digital sums additive? That is, for numbers A and B with digital sums "a" and "b", respectively, is it true that the digital sum "c" for C = A + B will be equal to the sum of "a" and "b"?

It may be here assumed (and has been elsewhere proved) that this is the case. That is, the digital sum of (9^62773 + 2)^83721 will be the digital sum of the sum of the digital sums: 9 + 8 = 17, 1 + 7 = 8.

So the answer should be "8".

Please check my work.

Eliz.

when did it come out?

So I got the same result in a simple way I made The counts (sorry if that's not the right word) and I got it the same way

lenny
It came out an hour ago, and it seems as though the answer is '8'. But don't blame me if it's not right, I've just been reading everything that was posted here.

thanks.i got 3rd place trophy just wondered if was right would i might get a higher one.

The calculation posted by stapel is correct. Just a couple of quick notes:
1. There is no need to assume that the power is odd. The calculations applied to the 9^x portion remain the same and we just need to look at a cycle of digital sums of powers of 2. That cycle is 6 long and is 2, 4, 8, 7, 5, 1, as stated in an earlier post. 83721 = 6 x 13953 + 3, so we need the third item from this cycle, or 8.



2. Not quite the case. c is actually the digital sum of a + b, which may not be equal to a+b, but rather need to be calculated as is done in the following line.

well I answered early and I hope to get the avatar and some trophie but I don't know if I can because this one was too easy
But I wish all of us the best luck

This was no where easy lol.

Hey I'm new here. This week's was pretty easy.

Never taken a course on number theory before, and brute forcing it is next to impossible, so you look for a pattern and check.

Seems 8 is indeed what you would get.

Note: It appears that you can post the math- or science-related Lenny's to this tutoring forum, and the tutors (math profs, etc) will give you the answers.

Eliz.

Yeah but they both got the answer wrong the first time.

Dear Holly, Congratulations! You have guessed correctly in the Lenny Conundrum game (round 234). You have won 478 NP! Because you were in the first 250 to guess correctly, you also have been awarded a Jazzmosis Grarrl Plushie, and receive a trophy and the Lenny Conundrum avatar! Yours Sincerely, The Neopets Team!

YAY! Thank you everybody. You are awesome.

Thanks!
Dear Jz,

Congratulations! You have guessed correctly in the Lenny Conundrum game (round 234). You have won 478 NP!

Because you were in the first 250 to guess correctly, you also have been awarded a Jazzmosis Grarrl Plushie, and receive a trophy and the Lenny Conundrum avatar!

Yours Sincerely,
The Neopets Team!

Thanks, guys! I'll try to contribute to the next conundrum. =)

lenny conundrum
What time does it usually come out?

The Lenny Archivist, always keeping an eye open for new job opportunities, decided to pay a visit to the Brightvale Institute of Technology and Engineering. Unfortunately, he showed up unannounced, and without an appointment, so the dean didn't have time to meet with him.

"Let me see if I have any free time," the dean said as he looked through the calendar. "But first, let me give you a little puzzle."

Then he wrote the following numbers on a piece of paper:

03 - 04 - 11 - 04 - 02 - 06 - 10 - 08

"Now, this code forms a word that would be part of your new job responsibilities. I don't think I could hire you unless you could solve it."

What was the word? Please submit only the one word answer. Any additional information will disqualify the entry
---
any ideas?

It can't be related to the numbers on the phone and doesn't appear to be related to position of letters in the alphabet. Can someone give some help, please?