Two questions to ask:

1. Does V(x) grow or decay?

2. When putting 4(0.65)^x vs 4(1.35)^x into the Y= of the TI-83/4, does the table look like the one above?

Hopefully the student will know that the number in parentheses (the growth/decay factor) determines if it is growth or decay but through the graphing calculator, they can figure it out, get the right answer and earn 2 points!

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# January 2018 Brain Puzzler Solution

Q: What is the sum of the digits of the square of ?

A: 81

Using the standard multiplication algorithm, whose digit sum is 81.

Or by looking at the pattern:

whose digit sum is 81

There is a shortcut to adding these digits

Reading from left to right, we can add the first 8 digits, 1 through 8 by making 4 pairs of 9 (1 + 8, 2 + 7, 3 + 6 and 4 + 5) making 36

Reading from right to left, we can add the last 8 digits (1- 8) the same way making 36.

Leaving the middle digit of 9.

36 + 36 + 9 = 81

# Get the Math and the Points! CC Alg I June 2017 Regents #10

Can use the TI-83/4 to figure this one out!

The zeros are the values of x that make f(x) or y equal to zero.

Answer (1) looks like this: only 1 zero at x = -3 which can be seen on both the graph and the table

Answer (2) looks like this with 3 zeros but only one of them is an integer value (both the table and the graph show that (-4,0) is a point

Let’s try answer (3):

Here we can see that on the table when x = -3 y = 0 and also when x = 0 and x = 4, y is also equal to zero. Looking at the graph we can see that the 3 x-intercepts are integer values of -3, 0 and 4. That’s it!

Pay close attention to the signs!! We need to look for the factors with the opposite sign to ‘zero it out’! This can be a bit counterintuitive as when -3 is a solution then the factor is (x + 3), as -3 makes x + 3 = 0. When 4 is a solution, the factor must be (x – 4) rather than (x+4) as plugging in 4 would not result in zero.

# December 2017 Brain Puzzler Solution

Q: For the positive integers up to and including 2017, how many of those integers have at least one zero?

A: There are positive integers in total to consider.

No one digit numbers have a 0 (1-9).

Below is the list of the qualifying numbers:

There are 9 two digit numbers: 10,20, 30…90 | 9 | |||||

There are the numbers 100 – 110 and 200-210 up to 900 – 910 | 99 | |||||

Also 120 130 140…190 and then 220 230 etc | 72 | |||||

Then 1000-1099 | 100 | |||||

Then same # of #s 1000 more than b | 99 | |||||

Then same # of #s 1000 more than c | 72 | |||||

2000-2017 | 18 | |||||

Total | 469 | |||||

Another solution is to exclude the numbers that do not have a zero and subtract from 2017:

9 one digit numbers

81 two digit numbers (9 x 9 : 9 choices for each digit 1-9)

729 three digit numbers (9 x 9 x 9: 9 choices for each digit 1-9)

For the numbers between 1000- 1999, there are the same amount as for the three digit numbers as they are the same numbers with a “1” in the thousands place value.

9 + 81 + 729 + 729 = 1548

2017 – 1548 = 469

# Get the Math and the Points! Common Core Algebra I Aug 2017 Regents #3

This can be easily solved by checking the answers: we will need a + and – to create a – so answers (1) and (2) are gone.

Look at (4): If we multiply (24.5x)(24.5x) we definitely get more than 49x^2 so finding this answer is easy.

Students can also put 49x^2 – 36 into Y1 on the TI-84 and then try each of the answers into Y2 and see which creates the same graph and/or table.

That’s it, 2 more points on the Regents!!