HP 35s User Manual

							 
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HP 35s  Base conversions and arithmetic 
 
hp calculators - 3 - HP 35s  Base conversions and arithmetic - Version 1.0 
Answer: 7640 base 8. Figure 2 shows the result in algebraic mode. 
 
Example 2: Add 7F6 base 16 to 1011001 base 2 and display the result in base 10. 
 
Solution: First, make sure the calculator is in HEX mode to enter the base 16 number.  
 
 $%+ 
  
 In RPN mode (note that  enters the hexadecimal digit F) 
 
 ,-$%-
$%(
&)&&))&$%.
$%&
/ 
 
 In algebraic mode: 
 
 ,#0-$%-
$%(
/
&)&&))&$%.
$%& 
 
 Figure 3 
 
 
 Figure 4 
 
Answer: 2127 base 10. Figure 3 shows the result in RPN mode. Figure 4 shows the result in algebraic mode. Note 
how entering the calculation as shown in algebraic mode makes use of the LASTx register. 
 
Example 3: Multiply FFF base 16 by 777 base 8 and display the result as a real number. 
 
Solution: First, make sure the calculator is in HEX mode to enter the base 16 number.  
 
 $%+ 
 
 In RPN mode: 
 
 $%-$%*
,,,$%,$%&1 
 In algebraic mode: 
 
 #0#0#0$%-$%*1
,,,$%,$%&  
						

							 
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HP 35s  Base conversions and arithmetic 
 
hp calculators - 4 - HP 35s  Base conversions and arithmetic - Version 1.0 
 
 Figure 5 
 
Answer: The result is 2,092,545. Figure 5 shows the result in algebraic mode. 
 
Example 4: Subtract 42 base 8 from 101111 base 2 and then display the twos complement of the result in base 2. 
 
Solution: First, make sure the calculator is in BIN mode to enter the base 2 number. 
 
 $%( 
 
 In RPN mode: 
 
 &)&&&&$%.$%*
(+$%,$%(23 
  
 In algebraic mode: 
 
 &)&&&&$%.$%*2
(+$%,$%(243
 
 Figure 6 
 
 Figure 7 
 
 Figure 8 
 
Answer: The result is 111111111111111111111111111111110011. This is equivalent to –13 in base 10.  
 
 Figure 9 
 
 Figures 6, 7, and 8 display portions of the answer in RPN mode.  
 
 Note that when the result is displayed, there is an arrow pointing to the left in the display to indicate that the 
result has digits in the answer that flow off the display screen to the left. Press the left cursor key one time 
to view the next 12 digits of the answer to the left and then press it again to view the leftmost 12 digits of 
the answer. Figure 9 shows the result converted into base 10. 
   
						

							 
 
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HP 35s  Using the LOGIC functions 
 
 
 
 
Numbers in different bases 
 
Operations on binary numbers  
 
Practice manipulating binary numbers 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
   
						

							 
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HP 35s  Using the LOGIC functions 
 
hp calculators - 2 - HP 35s  Using the LOGIC functions - Version 1.1 
Numbers in different bases 
 
Most numbers we work with day-to-day are in base 10. There are applications within the computer world that require the 
use of numbers in other bases. The number 24 in base 10 can be translated into base 16 by the following procedure. 
Just as each digit’s location in base 10 can be thought of as a power of ten (the ones’ place, the tens’ place, the 
hundreds’ place, etc), each digit’s location in base 16 can be thought of as a power of 16. Each digit in a base ten 
number can hold a value from 0 to 9. In base 16, each digit can hold a value from 0 to F, where F corresponds to the 
value 15 in a base 10 number. Translating 24 from base 10 to base 16 would require a 1 in the second location of the 
base 16 number (and would convert 16 of the 24 number’s value) and an 8 in the second location of the base 16 
number. Therefore, 24 base 10 is equal to 18 in base 16. A similar process could be used to convert 24 base 10 to base 
8 or base 2. 
 
On the HP 35s, numbers can be represented in bases 2, 8, 10 and 16, or binary, octal, decimal and hexadecimal. The 
HP 35s can work with numbers in bases 2, 8 and 16 that are 36 bits in length or less. Since the leftmost bit is used to 
indicate a negative number, the largest positive binary number is 0 followed by thirty-five 1s. This means that the largest 
hexadecimal number that can be entered or generated as an answer is 7FFFFFFFF, (equal to 34,359,738,367 in base 
10, and 377777777777 in base 8). This is because the HP 35s uses a 36 bit binary word space to represent numbers in 
these different bases. Decimal numbers are not limited in this fashion, since they can be represented as floating point 
numbers. The HP 35s calculator provides the ability to easily work with numbers in different bases, as the following 
sample problems illustrate. 
 
In RPN mode, the third row of keys on the HP 35s ( ! through  ) can be used to enter the hexadecimal digits A 
through F. In algebraic mode, it is necessary to press # and then the appropriate letter key to enter these digits. 
 
Operations on binary numbers  
  
Any binary number, regardless of the base in which it is displayed, can still be thought of as a collection of 1’s and 0’s. 
For example, the number 24 in base 10 is also the number 11000 in binary. This is because 11000 in binary is equal to 1 
x 24 + 1 x 23, or 24. The HP 35s contains many functions in the $> menu, shown below, that operate on binary 
numbers.  
 Figure 1 
 
 Figure 2 
 
The AND function compares two binary numbers at the bit-by-bit level and creates a new binary number with a 1 for 
each bit position that contains a 1 in both numbers in the same position.  
 
The XOR function (which stands for exclusive OR), does the same thing as the OR function, but only for positions where 
the original numbers contained a 1 and/or 0, but not where both contain a 1.  
 
The OR function compares two binary numbers at the bit-by-bit level and creates a new binary number with a 1 for each 
bit position if either original number contains a 1 at the same bit position.  
   
						

							 
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HP 35s  Using the LOGIC functions 
 
hp calculators - 3 - HP 35s  Using the LOGIC functions - Version 1.1 
The NOT function creates a new binary number where each bit position’s value has been “flipped”, with each 1 becoming 
a 0 and each 0 becoming a 1. This is referred to as the ones complement of the argument.  
 
The NAND function creates a new binary number from two input binary numbers where each bit position’s value is based 
upon a NOT (x AND y). In essence, it returns a bit of one unless both bit positions in the two input binary numbers are 
ones. 
 
The NOR function creates a new binary number from two input binary numbers where each bit position’s value is based 
upon a NOT (x OR y). In essence, it returns a bit of one ONLY when both of the bit positions in the two input binary 
numbers are zeroes. 
 
Note: Numbers entered into the HP 35s are assumed to be decimal numbers, regardless of the base mode, unless the 
proper suffix of “b”, “o”, or “h” is supplied. These suffix characters are found in the BASE menu as choices 5 through 8, 
and are shown below in Figure 1. It is not necessary to enter the “d” for decimal numbers, except for clarity in a program.  
 
 Figure 3 
 
Practice manipulating binary numbers 
 
Example 1:  Evaluate NOT(#4567 d). Make sure the HP 35s is in DEC mode to enter the base 10 number. 
 
Solution: In RPN mode, press: %!  then press#$%&$>$ 
 
 Figure 4 
 
 In algebraic mode, press: %!  then press $>$$%&& 
 
 Figure 5 
 
Answer: -4568 base 10. Figure 4 shows the result in RPN mode while figure 5 shows the result in algebraic mode. 
 
Example 2:  Perform an OR on these two binary numbers: #70114 o and #57610 o. Make sure the calculator is in 
OCTAL mode to enter the base 8 numbers. 
 
Solution: %!( 
  
 In RPN mode, press: 
#
#)$%!&%&)%!$>(#  
						

							 
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HP 35s  Using the LOGIC functions 
 
hp calculators - 4 - HP 35s  Using the LOGIC functions - Version 1.1 
 In algebraic mode, press: 
 
 $>(#
#)$%!%&)%!&#
 
 Figure 6 
 
 
 Figure 7 
 
Answer: 77714 base 8. Figure 6 shows the result in RPN mode. Figure 7 shows the result in algebraic mode.  
 
Example 3:  Perform an NOR on these two binary numbers: #1011 b and #1001 b. Make sure the calculator is in 
BINARY mode to enter the base 2 numbers. Do not forget to append the “b” suffix to the binary numbers. 
 
Solution: %!$ 
  
 In RPN mode, press: 
#
#)%!*&))%!*$>&#
 
 In algebraic mode, press: 
 
 $>&#
#)%!*))%!*&#
 
 Figure 8 
 
 Figure 9 
 
 Figure 10 
 
Answer: Figures 8, 9, and 10 show the result in algebraic mode. This display indicates an arrow pointing to the right 
which indicates that the answer scrolls off the screen in that direction. Press % to view the rest of the 
result. The leading 1’s in the answer are due to the 36 bit word length on the HP 35s. Since both binary 
numbers would have had all leading bits equal to zero, the NOR function returns a 1 for each of those zero 
bits. The last four bits reflect the NOR of the entered digits. Only the second entered bit was a zero in both 
numbers and NOR returns a 1 for that bit location.    
						

							 
 
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HP 35s  Using the formula solver – part 1 
 
 
 
 
What is a solution? 
 
Practice Example: Finding roots of polynomials 
 
Practice Example: Finding the root of a log equation 
 
Practice Example: Where there is no solution 
 
What the solver can and can not find 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
   
						

							 
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HP 35s  Using the formula solver – part 1 
 
hp calculators - 2 - HP 35s  Using the formula solver – part 1- Version 1.0 
What is a solution? 
 
Given a formula or equation such as: 
 
7x + 12 = 33 
 
it is easy to find that: 
 
x = 3 
 
This is the solution or root of the formula. It answers the question “What value of x makes the formula correct?” 
 
It is rarely this easy to solve a formula or equation. For example look at this one: 
 
4x² - 14LOG(x) = 33 
 
It seems likely that there is a value of x for which the formula is true, but there is no quick and easy way to find what this 
value is. A special method to find the solution is needed. 
 
The HP 35s has a Formula Solver built into it. This can be used to solve problems like the example above, and many 
others. This practice aid will show some simple applications. Part 2 will show further details.  
 
The Formula Solver is one part of the HP Solve application. Other practice aids show other uses of the HP Solver, for 
example its use to solve a formula typed as a program. 
 
Practice Example: Finding roots of polynomials 
 
Example 1: Solve the polynomial equation x³ + 15x² + 47x = -33 
 
Solution:  The HP 35s manual gives a program for finding the roots of polynomials up to the fifth order, but it is often 
simpler to use the formula solver. The following steps show how this can be done. 
 
 Before a solution is looked for, it is useful to store a guess in the variable to be used. In this case, the 
variable will be “X”, and to store a zero in it, the following leys are pressed: 
  
 !#$ 
  
 When # is pressed, the symbol A..Z is shown at the top of the screen as a signal that one of the 
keys marked A through Z at their bottom right must be pressed. To enter the variable X press the % key, 
which has a small “X” at its lower right to show that it is also the $ key. 
  
Next, the equation must be entered. This is done in the HP 35s “Equation mode.” Go to equation mode by 
typing &. Equation mode stores a list of equations and a new equation can be typed anywhere in this 
list. If necessary it is possible to put the new equation in a particular place by moving up or down through 
the list with the up and down cursor keys.   
						

							 
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HP 35s  Using the formula solver – part 1 
 
hp calculators - 3 - HP 35s  Using the formula solver – part 1- Version 1.0 
 Enter the equation by typing: 
  
 $()*%+,$(-*./,$012))3 
  
 To enter a variable into an equation, press the  key and then one of the letter keys. As with #, 
the symbol A..Z at the top of the screen is shown as a signal that one of the keys marked A through Z must 
be pressed. To enter the variable X in this formula press the % key again. 
  
 To solve the equation, press the 4 key. This is the lower left key below the display. The HP 35s 
asks which variable to solve for: 
  
 Figure 1 
  
 The symbol A..Z is at the top of the screen yet again. The variable in this formula is X so press the % key 
again to fill in X. 
  
 Figure 2 
  
 The word “SOLVING” is shown while a solution is looked for. If the search lasts more than a short time, the 
busy indicator 5 is shown as well. When a solution is found, it is displayed. It is also stored in the variable. 
  
 Figure 3 
  
Answer: X = -1 is a solution of the equation. 
 
The equation is a cubic, so there are two other solutions. If all three solutions are real numbers then the 
solution found depends on the initial guess given. Try finding a different solution. 
  
Example 2:  Solve the same polynomial equation with a guess of –50. 
 
Solution:  Store the new guess in the variable: 
  
 +!2#$ 
  
 Enter equation mode again by typing &. The equation is shown again. As equation mode stores a list 
of equations, it is possible to select a different equation to solve by moving up or down through the list with 
the up and down cursor keys below the screen. In this case the same equation is to be solved. 
    
						

							 
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HP 35s  Using the formula solver – part 1 
 
hp calculators - 4 - HP 35s  Using the formula solver – part 1- Version 1.0 
 Figure 4 
 
 To solve the equation, press 4, then press % again to show that the formula is to be solved for 
X. 
  
 Figure 5 
  
Answer: X = -11 is a second solution of the equation. There is a third solution, or root, which can be found if a 
suitable guess is tried. 
 
Practice Example: Finding the Root of a Log Equation  
 
Example 3: Polynomial equations of order up to 5 can also be solved using the polynomial solver program in the  
HP 35s manual, but many equations can only be solved using the function solver. Try solving the equation 
given at the beginning: 4x² - 14 LOG (x) = 33 
 
Solution: Enter the equation as before. Go to equation mode by typing &. Type the new equation immediately 
below the previous one, or move up or down through the list with the up and down cursor keys below the 
screen to place the new equation somewhere else in the list. 
  
 Enter the equation by typing: 
  
 .,6$78%.,09$701))3 
  
 This time, x² has been typed by use of the 6 key. Note that the function name is shown as SQ in 
equation mode: 
  
 Figure 6 
  
 To solve the equation, press the 4 key. Then press $ to show that this is the variable for which a 
solution is desired. A different variable name could have been used, but the same variable can be used in 
different equations. 
  
 The word “SOLVING” is shown as before. This time an error message follows. Although the solver knows 
which equation is being solved, it has started the search from the previous value stored in X, and that was 
–11, or one of the other negative roots of the polynomial in examples 1 and 2.