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Here are 3 introductory problems (AMC 8 Problems 1015), along with a CHALLENGE PROBLEM relating to the Foundations of NT. As always, each question adds 10 points to your leaderboard score, but the challenge problem adds 15 points!
1. What is the units digit of $2^{100}$?
2. Find the remainder when $2015^{2015}$ is divided by $14$?
3. If $p$, $q$, and $r$, are primes such that $7(p+q+r) = pqr$, then find $p$, $q$, and $r$.
CHALLENGE PROBLEM:
The twodigit integers form $19$ to $92$ are written consecutively to form the large integer
$N = 192021 · · · 909192$.
Suppose that $3^k$ is the highest power of $3$ that is a factor of $N$. What is $k$?
As always, you can post any questions you have in the Questions forum.
Surya DuraivenkateshGuest1.
The pattern of the units digit for powers of 2 is “2,4,8,6”, for example: 2^1=2, 2^2=4, 2^3=8, 2^4=16, 2^5=32, 2^6=64, etc.
So then we just follow the pattern till we get to 2^100 which will be with a units digit of 6.
adminKeymasterGood work Surya! You have been awarded 10 points! But, to further your understanding, how do you know that the “pattern” will continue. In other words, why doesn’t it go 2,4,8,6,2,4,8,6,3,5,…? How can we use modular arithmetic specifically to justify our answer more clearly?
Pratham S.GuestQuestion 2: We have the given equation. To keep our life easier, we would cancel out the 7. The best way to do that would be to assume one of the numbers say r, is equal to 7. Then we get pq = p + q + 7. We want to now factor this. You get pq – p – q = 7. Turn that into pq + p + q = 7. Then you get that (p1)(q1) 1 = 7, so (p1)(q1) = 8. The values of p and q have o be 3 and 5. So, p + q + r = 7 + 3 + 5(all are primes) = 15.

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