Carusi Middle School Problems of the Day

Week 17 hints and answers

As always, these hints and answers are intended to promote and provoke class discussions. The answers that are posted for students to read on the website a few days after each problem is posted are based in the answers that appear here.

ANSWERS TO THIS WEEK'S PROBLEMS

1. What is the sum of the prime factors of the number represented by:

2¹² - 2¹¹ +2¹⁰ - 2⁹ + ^{. . . . . .} +2² - 2¹?

For this one, let's start small and work back up to the entire sum, looking for a pattern. So

Sum	Value	Prime factors	Prime factor sum
2² - 2¹	4 - 2 = 2	2	2
2⁴ - 2³ + 2² - 2¹	16 - 8 + 4 - 2 = 10	2, 5	7
2⁶ - 2⁵ + 2⁴ - 2³ + 2² - 2¹	64 - 32 + 16 - 8 + 4 - 2 = 42	2, 3, 7	12
2⁸ - 2⁷ + 2⁶ - 2⁵ + 2⁴ - 2³ + 2² - 2¹	256 - 128 + 64 - 32 + 16 - 8 + 4 - 2 = 170	2, 5, 17	24
2¹⁰ - 2⁹ + 2⁸ - 2⁷ + 2⁶ - 2⁵ + 2⁴ - 2³ + 2² - 2¹	1024 - 512 + 256 - 128 + 64 - 32 + 16 - 8 + 4 - 2 = 682	2, 11, 31	44
2¹² - 2¹¹ + 2¹⁰ - 2⁹ + 2⁸ - 2⁷ + 2⁶ - 2⁵ + 2⁴ - 2³ + 2² - 2¹	4096 - 2048 + 1024 - 512 + 256 - 128 + 64 - 32 + 16 - 8 + 4 - 2 = 2730	2, 3, 5, 7, 13	30

There certainly doesn't seem to be any pattern here! But the answer to the question is 30.

2. There is definitely a pattern in the following triangle:

What will the 50th row of the triangle look like? Can you generalize your result for any row of the triangle?

This time, there really is a pattern. First of all, the number of numbers on the right side of the equals sign is 1 in the first row, 2 in the second, 3 in the third, and so forth. On the left side, there is one more number.
So in the 50th row, there will be 51 numbers on the left side and 50 numbers on the right.
And what are the numbers? Well, the first number on the left is always a perfect square. In fact, it is the square of the row number. So the 50th row should start with 50² = 2500, and it should have 51 numbers on the left side -- that is 2500 + 2501 + 2502 + ^{. . .} + 2550. And the next 50 numbers should be on the right: 2551 + 2552 + 2553 + ^{. . .} + 2600.
The sum on both sides is 128,775,
The general situation, for row n, is that there are n+1 terms on the left side, beginning from n² and ending with n² + n. On the right side there are n terms, starting from n² + n + 1 and ending at n² + 2n (notice that this is one less than (n + 1)² = n² + 2n + 1 (which will then be the first number on the next row).
For a special challenge, show that the sum on both sides of the nth row is n³ + 3n²/2 + n/2.

3. Fill in the missing terms of this pattern:

102, 105, 111, 114, 120, 123, 129, ____, ____, ____, ____, ____,..., ____, ____, 201, 204, 210, 213, 219, .....

This one's not too bad -- look at the differences between consecutive terms, and you'll see the pattern 3, 6, 3, 6, 3, 6....
So you alternately add 3 or 6 to get the next term. That means the filled-in sequence is:
102, 105, 111, 114, 120, 123, 129, 132, 138, 141, 147, 150, 156, 159, 165, 168, 174, 177, 183, 186, 192, 195, 201, 204, 210, 213, 219, 222, ...

4. - All of the positive integers are written in a triangular array as shown. In which row will the number 1000 appear?

Well, we don't want to write out the whole thing!
Is there a pattern we can take advantage of? Well, let's look at the last number in each row. 1, 3, 6, 10, 15, ...
These are the triangular numbers, and the one at the end of the nth row is half of n(n+1) -- for instance, 6 (at the end of the third row), is half of 3 times 4.
So we have to find out the first value of n for which n(n+1)/2 is bigger than 1000 (or what is the same, for which n(n+1) is bigger than 2000).
You've got to figure that the value of n we're looking for is pretty close to the square root of 2000, which is around 44.7.
So let's try 44^.45 = 1980 (too small), and then 45^.46 = 2070. This means that the last number in the 44th row is 990 (half of 1980) and the last number in the 45th row is 1035 (half of 2070, and 45 more than 990).
Therefore the number 1000 is (the 30th number in) row 45.

5. What is the sum of the numbers in the 100th row of the triangular array in yesterday's problem?

First off, we know from what we did yesterday that there are 100 numbers in row 100, and the last one is 100^.101/2 = 5050.
That means the sum of the numbers in row 100 is
4951 + 4952 + 4953 + ^{. . .} + 5048 + 5049 + 5050
We could add up these 100 numbers, or else we can use Gauss's trick of pairing the numbers from beginning and end, as follows:
(4951 + 5050) + (4952 + 4949) + (4953 + 4948) + ^{. .
.} + (5000 + 5001)
From this you can see that there are 50 pairs of numbers, and each pair adds up to 10,001. Therefore the sum of the entries in this row is 50 ^. 10,001 = 500,050.