Week 18

Example 1. Each element of the set {10, 11, 12, . . . , 19, 20} is multiplied by each element of the set {21, 22, 23, . . . , 29, 30}. If all these products are added, what is the resulting sum?

Example 2. In a building, there are 220 people. All of them either have a watch, a phone, or both. If 110 people have phones, and 20 have both a phone and a watch, how many people have only a watch?

Example 3. Suppose that f(x) = ax4bx2 + x + 5 and that f(−3) = 2. Find the value of f(3).

Of course, you can find the solution just below, but it is highly recommended that, you first try to solve it on your own.

Just remember the words of Paul Halmos, who says “the only way to learn mathematics is to do mathematics”.

Photo by Francesco Ungaro on Pexels.com

Solution 1. Initially, this is asking for the following product:

  • 10 󠄞× 21, 10 󠄞× 22, . . . , 10 󠄞× 29, 10 󠄞× 30;
  • 11 󠄞× 21, 11 󠄞× 22, . . . , 11 󠄞× 29, 11 󠄞× 30;
  • …             …             …
  • 19 󠄞× 21, 19 󠄞× 22, . . . , 19 󠄞× 29, 19 󠄞× 30;
  • 20 󠄞× 21, 20 󠄞× 22, . . . , 20 󠄞× 29, 20 󠄞× 30.

Next, the demand is to find the following sum

  • (10 󠄞× 21 + 10 󠄞× 22 + ··· + 10 󠄞× 29 + 10 󠄞× 30)
  • + (11 󠄞× 21 + 11 󠄞× 22 + ···  + 11 󠄞× 29 + 11 󠄞× 30)
  • + ···          ···              ···
  • + (19 󠄞× 21 + 19 󠄞× 22 + ···  + 19 󠄞× 29 + 19 󠄞× 30)
  • + (20 󠄞× 21 + 20 󠄞× 22 + ···  + 20 󠄞× 29 + 20 󠄞× 30).

Finally, note how we can write each term of the above row.

  • (10 󠄞× 21 + 10 󠄞× 22 + ··· + 10 󠄞× 29 + 10 󠄞× 30) = 10(21 + 22 + ··· + 29 + 30)
  • (11 󠄞× 21 + 11 󠄞× 22 + ···  + 11 󠄞× 29 + 11 󠄞× 30) = 11(21 + 22 + ···  + 29 + 30)
  • …             …             …
  • (19 󠄞× 21 + 19 󠄞× 22 + ···  + 19 󠄞× 29 + 19 󠄞× 30) = 19(21 + 22 + ···  + 29 + 30)
  • (20 󠄞× 21 + 20 󠄞× 22 + ···  + 20 󠄞× 29 + 20 󠄞× 30) = 20(21 + 22 + ···  + 29 + 30).

But 10 + 11 + ··· + 20 = 165and21 + 22 + ··· + 30 = 255.

Therefore, required sum is

10(21 + 22 + ··· + 29 + 30) + 11(21 + 22 + ···  + 29 + 30) + ···+19(21 + 22 + ···  + 29 + 30) + 20(21 + 22 + ···  + 29 + 30)

= (10 + 11 + ··· + 19 + 20) (21 + 22 + ··· + 29 + 30)

= (165)(255)

= 42075.

Solution 2. There are two different sets in this problem: the set of all the people with a watch and the set of all the people with a phone.

The phone set has 110 elements, the intersection of the sets has 20 elements, and the union of the sets has 220 elements.

Let us call the number of people with watches w, and the number of people with phones p. We already know that p = 110, so we need the value of w.

From the set analysis we did before, we know that 110 + w − 20 = 220. Therefore, w = 130.

Here is a Venn Diagram that represents the scenario.

From this we can easily see that w – 20 + 20 + 90 = 220, which yields w = 130.

Solution 3. We first need to relate the value of f(−x) to the value of f(x). We then apply this relationship when x = 3. Notice that

(−x)n = xn when n is an even integer, and

(−x)n = −xn when n is an odd integer.

Applying this information implies that

f(−x) = a(−x)4b(−x)2 + (−x) + 5

= ax4bx2x + 5

= (ax4bx2 + x + 5) − 2x

= f(x) − 2x.

Therefore, f(3) = f[−(3)] − 2(−3) = 2 + 6 = 8.

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