I've just posted the eighth Problem Of the Week, along with a solution to last week's problem. Only two more problems to go before we hang up our spurs until the fall.
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I have just posted the penultimate POTW for the term, along with the “official” solution to last week's problem. Only one more problem after this, then it's nothing until the fall. Enjoy them while they last!
I've just posted the new Problem of the Week, along with an official solution to last week's problem. But this one will have to hold you for a while, since I'm taking next week off.
The second Problem of the Week has now been posted. More of a puzzle this week, rather than a conventional math problem. Enjoy! I've also posted a solution to last week's problem. Enjoy that too!
Gotta run now. Breaking Bad is on in just over two hours, and I have to begin my preparations.
My recent travels, to Parsippany, NJ via Baltimore, MD, which involved three talks in two days, followed by multiple games of chess, bookended by two long drives, came to a dramatic conlcusion yesterday when I had to drive home in the snow. Not fun! There was so much snow on the road that you…
If I understand correctly: multiply 10 by each element of B, then 11 by each element of B, ..., and the sum of those:
There may be a shorter method but:
write each element of A as
ai = 10 + i, 0 <= i <=10
and for B
bj = 20 + j, 1 <= j <= 10
write xij = ai * bj
Then set xi* = sum over j of xij
(use formulas for sum of integers)
Then answer is sum over i of xi* (formula for sum of integers used again). The answer I obtained (this way and using R by "brute force" as check) is surprisingly large.
As I understand what is being asked, and to simplify to a more reasonable sized problem, there are two sets:
If the first set is {a,b,c,d} and the second is {A,B,C} what is the sum =
a*A + a*B + a*C +
b*A + b*B + b*C +
c*A + c*B + c*C +
d*A + d*B + d*C
Which simplifies to
A*(a+b+c+d) +
B*(a+b+c+d) +
C*(a+b+c+d)
And further to
(A+B+C)*(a+b+c+d)
Going back to the original larger problem
(10+11+12+13+14+15+16+17+18+19+20)*
(21+22+23+24+25+26+27+28+29+30)
The first sum = 11*(10+20)/2 = 11*15 = 165
The second sum = 10*(21+30) = 5*51 = 255
So the total = 165 * 255 = 42075
I checked with the following:
sum = 0
for i = 21,30 do
for j = 10,20 do
Sum = sum + i*j
end
end
print(sum)
However, I see that dean above had a different understanding of the problem,which would indeed produce a surprisingly large number.
I see I forgot to type a "/2" in the second sum line. It should be 10*"(21+30)/2" I did the divide-by-two after the next equals sign.
actually RickR, i obtained the same value you did. Here is my R code:
x<-10:20
y<-21:30
xy<-x%o%y
sum(xy)
returns 42,075
I posted this question on the previous POTW thread:
====================================
Let A be the set {10, 11, 12, . . . , 20}, and let B
be the set {21, 22, 23, . . . , 30}. Each element
of the first set is multiplied, in turn, by each
element of the second set. Find the sum of all
these products.
====================================
Does “in turn…by each” mean A1*B1+A2*B2+A3*B3+….+A10*B10
or
A1*(B1+B2+B3+…+B10)+A2*(B1+B2+B3+…+B10)+….+A10*(B1+B2+B3+…+B10)
??
I think the latter is probably right, but I'm not sure.
RickR's understanding of the problem in comment two is correct, and his explanation is precisely the one I include in the official solution.
Hmm. I am wondering why this isn't trivial.
In effect, this can be rewritten as:
10(21) + 10(22) + ... 10(30)
+ 11(21) 11(22) + ... 11(30
...
Which is equivalent to:
10(21 + 22 + ... + 30)
+ 11(21 + 22 + ... + 30)
...
Which is in turn equivalent to:
(10 + 11 + ... + 20)(21 + 22 + ... + 30)
So all you really need to do is a couple of quick sums.
10 + 11 + ... + 20 = (10 + 20) + (11 + 19) + ... (14 + 16) + 15 = 30 * 5 + 15 = 165
Similarly, 21 + ... + 30 = 51 * 5 = 255
So it's just 165 * 255 = 42075, which as several posters above me have confirmed is correct.