POTW 8 Posted

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.

More like this

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.

By Another Matt (not verified) on 28 Mar 2016 #permalink

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.