Questions and Answers #8 Decimals

Questions and Answers Sheet 8 Decimals Question #1 Finding the smallest n such that 𝑛2 ends with 00001 The problem is to find the smallest natural number n so that its square's last 5 digits are 00001. n and 𝑛2 cannot begin with 0. Answer: You want to find n such that n2  1 ( mod10 ) 5 By Chinese remainder theorem, this is equivalent to ( mod 2 ) and n2  1 5 ( mod 5 ) n2  1 5 We have ( mod 2 )  n  1 ( mod 2 ) n2  1 5 4 and ( mod 5 )  n  1 ( mod 5 ). n2  1 5 5 Again by Chinese remainder theorem, these congruence relations give four possibilities of 𝑛 𝑚𝑜𝑑 5 ⋅ 104 , namely n  1,18751,31249, 49999 ( mod 50000) . From here, we see that the smallest such n is 18751. Question #2 For questions like "What is the Answer: Step 1 The definition is: 𝑎 1 √𝑥 = 𝑥 𝑎 . 1 Hence, if 𝑎 = , then: 𝑏 1 𝑏 √𝑥 = 𝑥 𝑏 . As a consequence: 1 2 th root of x would the answer be 𝑥 2 ? 1 2 √𝑥 = 𝑥 2 . Step 2 Notice that 𝑎 = 1 𝑏 implies that, if for example 𝑏 = 22 91 , then 𝑎 = 91 22 . If for example 𝑏 = √3, then 𝑎 = 1 √3 . And so on. Question #3 Function to grab specific digits from a number? Recently I’ve been thinking a lot about grabbing digits from numbers, for things like calculating multiplication persistence, i.e. turning 1234 into a 1 × 2 × 2 × 3 × 4 . I’ve been able to come up with [log 𝑥] + 1 to output the number of digits of the given number x, as well as x   10logx −n +1  which outputs the first n digits of any number x. But that’s about as far as I’ve been able to think up. Is there any way to construct a function that retrieves the nth digit of a given number? Forgive me if my notation or comprehension is poor, I have no formal education in this field of maths. Answer: If the digit you require is n places from the right (indexed from zero so the units place is at position 0, the tens place is at 1 and so forth), then the following function will return the n-th digit of the decimal integer x:  x  f ( x, n) =  n  10  ( mod10 ) Example: f(123456, 3) = 3 You can replace 10 everywhere in the formula with any base b to generalise it. Question #4 Given a natural number 𝑛 < 109 , find the maximum number of the multiple of 3, which differs by exactly one digit from the given one Answer: Step 1 𝑘 Let 𝑛 = ∑𝑠−1 𝑘=0 10 𝑑𝑘 be an s-digit number when written in base 10. s −1 Let tc = ( d k ) mod 3 be the sum, modulo 3, of all the digits except the c'th one. k =0 k c The algorithm goes: Set 𝑐 = 𝑠 − 1 Step 2 If digit 𝑑𝑐 < 9 − 𝑡𝑐 , replace digit 𝑑𝑐 with 9 − 𝑡𝑐 and exit Step 3 Set c=c-1 If 𝑐 ≥ 0 return to Step 2. If c<0 exit: there is no solution for this n. A few examples follow. 𝑛 = 523, 𝑡2 = 2, 𝑡1 = 2, 𝑡0 Step 2 (c=2) replaces the 5 with 9-2=7 and exits with 723 𝑛 = 826, 𝑡2 = 2, 𝑡1 = 2, 𝑡0 = 1 Step 2 (c=2) decides that the 8 is too big to be replaced. Step 2 (c=1) replaces the 2 with 9-2=7 and exits with 876. 𝑛 = 879, 𝑡2 = 1, 𝑡1 = 2, 𝑡0 = 0 . Step 2 decides that all the digits are too big to be replaced. (8 ≥ 8,7 ≥ 7,9 ≥ 9). There is no solution for 879. Question #5 How do you change 7 8 to a decimal and percent? Answer: You could long divide, but let's try a different approach. Since 2 is a factor of 10, we have 8 = 2 × 2 × 2 is a factor of 1000 and 1000 = 5 × 5 × 5 × 5 = 125 8 Then: 7 8 = 7×125 8×125 = 875 1000 = 0.875 Question #6 How do you convert 3 50 to decimal and percentage? Answer: A decimal is any fraction which denominator is a power of 10. So to convert a fraction to a decimal you look for a number to which you can expand or reduce the fraction. 3 50 3 50 can easily be converted to a fraction with the denominator of 100 = 3×2 50×2 = 6 100 This fraction can easily be written as a decimal or percentage: 6 100 = 0.06 = 6% Question #7 How do you convert 2\frac{2}{5} into a percent and decimal? Answer: The number 1 is the same as 1.00 and is the same as 100%. The number 2, therefore, is the same as 2.00 and the same as 200% 2 5 can be converted into a decimal by seeing that 2 5 = 4 10 × 4 10 , spoken out loud, is "four tenths" and shows that the decimal is 0.4. And 0.4 is the same as 40% . 2 Which means that 2 = 2.4 = 240% 5 Question #8 3 How do you convert 3 into a decimal and percent? 4 Answer: To turn 3 3 4 to a decimal, you must first convert it to an improper fraction, to do that you multiply the whole number (3) and the denominator (4) 4 × 3 = 12 Then you add the product of 3 and 4(12) to the numerator (3) 3+12=15 You then make the sum of the numerator and the product of the whole number and denominator into the numerator. This leaves us with 15 3 Now, to make that into a percent is easy. 15 3 × 𝑥 100 You need the multiply 15 and 100 15 × 100 = 1500 Then you divide the product by 3 1500 = 500 3 The percent is 500% Question #9 How do you convert 6 15 25 into a percent and decimal? Answer: To convert a fraction in decimal, one should make denominator a power of 10. Here as denominator is 25, we can easily do it by multiplying numerator and denominator by 4. Hence 6 =6 6 10 15 =6 25 =6+ 6 10 15×4 25×4 + =6 0 100 60 100 = 6.60 And to convert a fraction in percent, we have to just multiply it by 100. Hence 6 15 25 = 6.60 = 6.60 × 100% = 660% Question #10 How do you evaluate the following expression using decimals? 1 2 5 3 4 25 − + Answer: 1 2 5 3 4 25 − + = = 1 × 50 5 × 25 3 × 4 − + 2 × 50 4 × 25 25 × 4 50 100 − 125 100 + 12 100 = 0.50 − 1.25 + 0.12 = −0.63 Question #11 How do you evaluate the following expression using decimals? 4.136 − 8.2612 + 77 64 + 205 128 Answer: Here 77 64 and 205 128 have only 2 as ' factors for denominators. In fact they are 26 and 27 and hence they can be converted into decimal form by multiplying numerator and denominator by 56 = 15625 and 57 = 78125 Hence 77 64 = 1.2031125 and 205 128 = 1.6015625 Hence 4.136 − 8.2612 + 1.203125 + 1.6015625 = −1.3205125 Question #12 How do you evaluate the following expression using decimals?: 4 17 5 32 4.56 − + 3.21 + Answer: The best way is to convert each fraction to decimal form. Two of them are already in decimal form 4 5 = 4×2 5×2 = 8 10 = 0.8 and 17 32 = 17×3125 32×3125 = 50125 100000 = 0.50125 Hence, the expression becomes 4.56-0.8+3.21+0.50125=7.47125 Question #13 How many digits does 225964951 − 1 have? Answer: Taking the 10-logarithm is basically the easiest way to do it. Remember to always round down, no matter what the decimal part is (for instance, log10 (99) + 1 ≈ 2.996 , and we want the answer to be 2). However, if all you have is a classic calculator that won't work with such large numbers, or you want to get an approximate answer with just mental calculations, some more manipulations must be done first. Start with noting that 225,964,951 − 1 and 225 964 951 have the same number of digits. From there, we know that log10 (225 964 951) = 25 964 951 ⋅ log10 (2) And then either let the calculator do the work, or if you go the mental route, know that log10 (2) ≈ 0.3(𝑓𝑟𝑜𝑚210 ≈ 103 ) and just get 25 964 951 ⋅ log10 (2) ≈ 7 800 000 For the mental calculation, adding 1 here (although it's technically what you're supposed to do) makes no sense in my opinion. Question #14 How to demonstrate the method used to convert decimal to any other base? Let's say we have an arbitrary number in base b, (𝑥3 𝑥2 𝑥1 𝑥0 )𝑏 . Answer: You misinterpreted the video. 𝑁 𝑏 = 𝑥3 × 𝑏 2 + 𝑥2 × 𝑏1 + 𝑥1 × 𝑏 0 with a remainder of 𝑥0 To take a simpler example, consider dividing 14 by 3. Then 14 3 = 4 , with a remainder of 2. This implies that 14 = (3 × 4) + 2 Question #15 How to formally define decimal truncation? f n (r ) = 10− n sign(r ) 10n | r | Answer: f n (r ) = 10− n sign(r ) 10n | r | Where  x  is the greatest integer ≤ 𝑥 This can be written as: 1) 𝑓𝑛 (𝑟) = 10−𝑛 𝑓0 (10𝑛 𝑟) The use of absolute values and sign reflects that it is somewhat arbitrary to “truncate” towards zero. For example, if r is positive and n is a positive integer, then 𝑓0 (𝑟 + 𝑛) = 𝑓0 (𝑟) + 𝑛, but this doesn’t generalize to all real r and integers n. You could substitute  1 g0 (r ) =  r +   2 and define 𝑔𝑛 using the equivalent of (1) to define rounding to n digits. Rounding is more natural because 𝑔0 (𝑟 + 𝑛) = 𝑔0 (𝑟) + 𝑛 for all real r and integer n. The only time you will get 𝑓0 (−𝑟) ≠ −𝑓0 (𝑟) is when r is exactly a half-integer. Question #16 How to make my algorithm work for case: irrational number √2 so that | p − 2 | 0.001 q Answer: Your algorithm is applicable to approximate rationals by rationals, but it doesn’t applicable directly to approximate reals by rationals. On the other hand, best rational approximations of a given a real number are provided by its simple continued fraction. Moreover, for rational approximants, listed here, we see that | 41 − 2 | 4.21 10 −4 29 Question #17 How to multiply 2 Binary Numbers? a = a N a N a1 , 2 2 b = bN bN b1 −1 2 2 −1 Answer: In base 10, the digit 𝑎1 can take values from 0 to 9. Any numbers higher than that can only displayed if you also use 𝑎2 . In your setting, 𝑎1 can take values from 0 to 65535 (which is 1111111111111111 in binary). Any numbers higher than that can only displayed if you also use 𝑎2 . So, in summary, we should consider base 65536 (instead of base 10). The formula becomes a  b = a N a N  a  bN bN  2 N 2 2 −1 2 2 −1 N 2 b1 =  (ai b j )65536i + j − 2 , i =1 j =1 where instead of 65536𝑖+𝑗−2 we can also write (216 )𝑖+𝑗−2 , which is the same as 216(𝑖+𝑗−2) . The number 16(𝑖 + 𝑗 − 2) in the exponent of the last expressions is the number of bits that you need to shift by. to do (because if you multiply by 2𝑘 this corresponds to a shift by k bits). Question #18 How would you graph − 5 4 on a number line? Answer: A number line extends from −∞ on the extreme left to ∞ on the extreme right. Zero will be in the center. Just one unit to the right of zero is 1 and -1 one unit too the left of zero. 5 1 4 4 − = −1 , hence it lies between -2 and -1, -2 will be one unit further to the left of -2 by one unit. If we divide the distance between -2 and -1 in to four equal parts , −1 1 4 will be the one just left of -1 (or three fourths to the right of -2). Question #19 If 9.999 ⋯ = 10 , then is there a general proof for any number that has infinite trailing 9s? Answer: First of all, assume that 𝑥 ∈ (0,1) . Then 𝑥 = 0. 𝑑1 𝑑2 … 𝑑𝑛 99999 … , with 𝑑𝑛 ≠ 9 . Therefore 10𝑛 𝑥 = 𝑎 + 0.999 …, where 𝑎 = 10𝑛−1 𝑎1 + 10𝑛−2 𝑎2 + ⋯ + 𝑎0 So, 10𝑛 𝑥 = 𝑎 + 1 , and therefore 𝑥 = 0. 𝑑1 𝑑2 … 𝑑𝑛−1 (𝑑𝑛 + 1) In the general case, x can be written as m+x' , with 𝑚 ∈ 𝑍 and where 𝑥 ′ ∈ (0, 1) has trailing 9's. So, the previous argument aples to x' Question #20 Infinite binary having infinite decimal 1 2 = 0.5 = (0.1)2 Answer: We have 1 2 1 3 1 5 = 0.5 = (0.1)2 = 0.333 ⋯ = (0.010101 ⋯ )2 = 0.2 = (0.001100110011 ⋯ )2 So it can happen that both expansions are finite both are infinite the decimal is finite and the binary is infinite. What cannot happen is that the decimal is infinite and the binary is finite. May be that's what you are trying to prove. In other words we have the implication: Decimal infinite ⇒ decimal finite. To prove this observe that if 𝑥 = (0. 𝑏1 𝑏2 … 𝑏𝑛 )2 then 𝑥 = 𝑚 2𝑛 = 𝑚 5𝑛 10𝑛 which has a finite decimal expansion.