Questions and Answers #7 Decimals

Questions and Answers Sheet 7 Decimals Question #1 A person invests $2500 in an account at 6% interest compounded annually. Let V=f(t) be the value (in dollars) of the account after t years or any fraction thereof. a. Find an equation of f. b. What is the V-intercept? What does it mean in this situation? c. What will be the value of the investment in six years? d. When will the value of the investment be $7000? Answer: Given that P=$2500 time =t years rate =6% Principal rate =$2500 Rate =6% compomoted annualy time =t (general) using formula 𝜕 𝑛𝑡 𝑣 = 𝑓(𝑡) − 𝑝 (1 + ) 𝑛 ⇒ 2500 (1 + 6 100 ) 𝑡 A) [𝑣 = 𝑓(𝑡) ⇒ 2500(1.06)𝑡 ] B) The v-intercept (x,y)=(0,2500) This indicate the original investment was 2500 Dollars C) at t=6 year 𝑣 = 2500(1.06)6 v=3546.297 (Dollars) D) The time at which value will $7000. 7000 = 2500(1.06)𝑡 70 = (1.06)𝑡 25 Taking log both sides 14 1.029619 ln = 𝑡 ln(1.06) ⇒ 𝑡 = 5 0.58268 t=17.670 years Question #2 How wide is the chasm between what men and women earn in the workplace? According to a 2015 analysis from a national women's group, women lose $435,049 over the course of a career because of the pay gap. The bar graph to the right shows the average earnings in the United States for men and women at ages 24 and 60. This exercise involves the graphs of models for the data shown in the rectangular coordinate system to the right. Complete parts (a) through (c) below. a. Use the two points for men shown on the graph to find a function in the form M(x)=mx+b that models average yearly earnings for men x years after age 24. M(x)=? Answer: Find the equation for men using two given points two points are (𝑥1 , 𝑦1 ) = (0,25) … y-intercept is b=25 (𝑥2 , 𝑦2 ) = (30,64) the slope of the line is given by 𝑦 −𝑦 𝑚= 2 1 𝑚= 𝑚= 𝑥2 −𝑥1 64−25 30−0 13 10 𝑚 = 1.3 Equation of the line is given by y=mx+b y=1.3x+25 M(x)=1.3x+25 answer(a) NOTE: Refresh your page if you can't see any equations. Question #3 Solve a) 7.456 × 8 b) 56 × 35 Answer: Step 1 a) 7.456 × 8 = 7.456 8 59.648 ∴ 7.456 × 8 = 59.648 Step 2 b) 56 × 35 = 1960 56 35 280 168 1960 Question #4 What is 1 3 as a percent? Answer: Convert it into a decimal 1 10 = 0.1 Change a decimal to a percentage by multiplying by 100 0.1=10% 1 so = 10% 10 Question #5 Bank Account A has $750.92, and Bank Account B $675.44. Account A changes by -$216.38, and Account B changes by -$168.49. Which account has the greater balance? Answer: Bank Account A has a balance of $750.92 and it changes by -$216.38. Hence, its balance becomes 750.92 − 216.38 = $534.54 . Bank Account B has a balance of $675.44 and it changes by -$168.49. Hence, its balance becomes 675.44 − 168.49 = $506.95 . Hence Bank Account A has greater balance. Question #6 A periodic decimal expansion sequence Let (𝑥𝑛 )𝑛∈𝑁 be a sequence with the following terms. 𝑥1 = 0, 1 𝑥2 = 0, 13 𝑥3 = 0, 131 𝑥4 = 0, 1313 𝑥5 = 0, 13131 and son on. Prove lim 𝑥𝑛 = 𝑛→∞ 13 99 Answer: I'll start with an independent proof that . . 13 = 0.13 : 99 Let . . x = 0.13. Then . . 100 x = 13.1399 x = 100 x − x . . . . = 13.13− 0.13 = 13 x . . 13 = = 0.13. 99 . . 13 = 0.13 = 0.13131313 . 99 Hence for all natural numbers k, we have: 13 1 0 < − 𝑥𝑘 < 𝑘. 99 10 Take limits as 𝑘 → ∞ of the above inequality and quote the Squeeze Theorem. An alternative method would be to prove that the sequence is Cauchy which is similar to the above method in fact, and then use the fact that every real Cauchy sequence (with the usual metric) is convergent. Question #7 Why sum of digits in even base have such property? Function D(a, b) define as sum of digit of a in base b. Example D(5,2)=2 . 𝐷(𝑘,𝑛) Let 𝑓(𝑚, 𝑛) = ∑𝑚 𝑘=1(−1) Answer: The point is that if l, 𝑛 are even then (−1)𝐷(l,𝑛) + (−1)𝐷(l+1,𝑛) = 0 This is because \ell and l + 1 differ only in the last digit in base n. 𝐷(𝑘,𝑛) That means ∑𝑙+1 = 0 for l even, because it can be split into l + 1 pairs which each sum to 0. 𝑘=0(−1) Now if 𝑚 = l + 1 is odd then 𝐷(𝑘,𝑛) 𝑓(𝑚, 𝑛) = ∑l+1 − (−1)𝐷(0,𝑛) = 0 − 1 = −1, 𝑘=0(−1) whereas if 𝑚 = l is even then 𝐷(𝑘,𝑛) 𝑓(𝑚, 𝑛) = ∑l+1 − (−1)𝐷(0,𝑛) − (−1)𝐷(l+1,𝑛) = −1 − (−1)𝐷(l+1,𝑛) , 𝑘=0(−1) 𝐷(l+1,𝑛) and since (−1) ∈ {−1,1} , we get 𝑓(𝑚, 𝑛) ∈ {−2,0} Question #8 1 2 −(𝑛/10) Why is ∑∞ almost equal to 5√𝜋 − (agreeing up to 427 digits)? 𝑛=1 𝑒 2 Answer: Define 1 1 2 𝜓(𝑞) ≡ − + 𝜗3 (0, 𝑒 −𝜋𝑞 ) = ∑ 𝑒 −𝜋𝑛 𝑞 2 2 𝑛≥1 where 𝜗3 (𝑧, 𝑞) is Jacobi's third theta function. An application of the Poisson summation formula leads to the identity 𝜗3 (0, 𝑒 −𝜋/𝑞 ) = √𝑞 𝜗3 (0, 𝑒 −𝜋𝑞 ) whose proof is outlined in Jacobi (1828). Substituting 𝑞 ≔ (𝜋𝑘 2 )−1 yields 2 2 ∑𝑛≥1 𝑒 −𝑛 /𝑘 2 1 1 = − + 𝜗3 (0, 𝑒 −1/𝑘 ) 2 1 2 1 2 2 = − + ⋅ √𝜋𝑘 2 𝜗3 (0, 𝑒 −𝜋 2𝑘2 ) so that 2 2 ∑𝑛≥1 𝑒 −𝑛 /𝑘 1 𝑘 1 𝑘 = − + √𝜋(1 + 2𝜓(𝜋𝑘 2 )) 2 2 2 = − + √𝜋 + 𝑘√𝜋 ∑𝑛≥1 𝑒 −(𝜋𝑘𝑛) . 2 2 When k=10 the error has an order of magnitude of 2 log10 (10√𝜋𝑒 −100𝜋 ) = 1 + log10√𝜋 − 100𝜋 2 log10 𝑒 = 𝑓 − 427 which corresponds to your observation. Question #9 Why are repeating decimals often self-inversions? Given a reptend r with 2k digits as in 1 = 0. ⋯ 𝑟1 𝑟2 ⋯ 𝑟(2𝑘) , 𝑥 you will often find that 𝑟𝑖 + 𝑟(𝑘+𝑖) ≡ −1(𝑚𝑜𝑑 𝑏) for all 𝑖 ≤ 𝑘 Answer: That the decimal expansion of 1 𝑝 repeats after 2k digits means 𝑏 2𝑘 ≡ 1 (𝑚𝑜𝑑 𝑝) which means 𝑏 𝑘 ≡ ±1(𝑚𝑜𝑑 𝑝) and if it doesn't repeat after k digits that means 𝑏 𝑘 ≡ 𝑝 − 1 𝑚𝑜𝑑 𝑝 1 Thus, if = 0. 𝑟1 𝑟2 … 𝑟2𝑘 , then 𝑝 (𝑟1 𝑟2 ⋯ 𝑟𝑘 ) × 𝑝 = 𝑏 𝑘 − (𝑝 − 1) and (𝑟𝑘+1 𝑟𝑘+2 ⋯ 𝑟2𝑘 ) × 𝑝 = (𝑝 − 1) × 𝑏 𝑘 − 1, so (𝑟1 𝑟2 ⋯ 𝑟𝑘 + 𝑟𝑘+1 𝑟𝑘+2 ⋯ 𝑟2𝑘 ) × 𝑝 = 𝑝 × (𝑏 𝑘 − 1) i.e., 𝑟1 𝑟2 ⋯ 𝑟𝑘 + 𝑟𝑘+1 𝑟𝑘+2 ⋯ 𝑟2𝑘 = 𝑏 𝑘 − 1 Question #10 Where to stop when doing decimal division? We know that 1 8 = 0.125 via calculator; however, if I didn't have access to a calculator and wanted to find this via long division, why would I stop at 3 decimal places? Why not 2 or 4? Answer: If I understood your question, you are searching for a method to make a division without a calculator. 1 Similar to Miscellaneous answer, in the case of this is what I would do: 8 You can write the rational like this: 1 = 8 1 2⋅2⋅2 = 1 2 2 2 This means that you need to take the mid part of a number three times, so 1 = 0.5 2 0.5 2 = 0.25 0.25 2 = 0.125 Question #11 Where does 3 4 lie on the real 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 to the left of zero. 3 4 lies between 0 on the left and 1 on the right. Divide this portion in four equal parts. 3 4 will be one part to the left of 1. Question #12 When is 𝑋 𝑅 (𝑋) an integer where R(X) is the reverse of an integer X? Answer: Step 1 There are no solutions with a ratio of 2 or 3. In fact the only possible ratios are indeed 4 and 9. Let x be the smaller number with terminal digit a, and y be thevlarger number either terminal digit b. Then 𝑦 y has more digits than x, which kills us, unless the ratio = 𝑟 has only one digit. 𝑥 Trying r=2 , we observe that in the units place we must have 2𝑎 ≡ 𝑏 (𝑚𝑜𝑑 10) , and the leading digit must satisfy one of the following: 2b=a 2b+1=a Pairing 2a  mod10 with 2b=a implies 4𝑎 ≡ 𝑎 b 𝑚𝑜𝑑 10 forcing a=0 , which is no good (the product should not have a leading digit of 0). Pairing 2𝑎 ≡ 𝑏 𝑚𝑜𝑑 10 with 2b=a gives 4𝑎 + 1 ≡ 𝑎 𝑚𝑜𝑑 10 from which we can try a=3 , but then 2 × 6 + 1 = 13 ≠ 3 (we must have exactly 3, not just 3 mod 10, or else y has too many digits). Similar logic eliminates 3 as a possible ratio. For ratios from 5 to 8, we use the fact that b, which must be the leading digit of x as well as the last digit of y, can only be 1. Thus 𝑟/∈ {5,6,8} as these fail to be units mod 10. For r=7 we can try a=3 , but then the leading digit of y would have to be 3 whereas the actual product 7x, if it has the same digits as x, has to begin with 7,8, or 9.as these fail to be units mod 10. For r=7 we can try a=3 , but then the leading digit of y would have to be 3 whereas the actual product 7x, if it has the same digits as x, has to begin with 7,8, or 9. So we are left with the ratios that are well known to have solutions, 4 and 9. In both cases there are infinitely many solutions. The minimal ones for each ratio are of course 1089 × 9 = 9801 2178 × 4 = 8712 In each case we may split the first two digits from the last two and insert an arbitrary number of 9's, as in 1099989 × 9 = 9899901 2199978 × 4 = 8799912 with three 9's inserted (these being the only solutions, by the way, with seven digits). There are also solutions like the following: 1099109999899989 × 9 = 9899989999019901 2199219999789978 × 4 = 8799879999129912 Step 2 Solutions do exist with ratios of 2 or 3 in other bases. Among them: 1 − 12 × 2 = 2101 base three 1023 × 3 = 3201 base four 13 × 2 = 31 base five 4378 × 2 = 8734 base twelve 3289 × 3 = 9823 base twelve Question #13 When does the fraction 142 2.54 ever repeat? Answer: Note that 142 2.54 = 100 × 71 127 Since 127 is ', it may well happen that the decimal expansion of 71 127 is periodic with period 126. But that's not the case. It turns out that it is periodic with period 42 (= 126 3 ) 42 digits 71 127 = 0. ⏞ 559 055 118 110 236 220 472 440 944 881 889 763 779 527 … and then it repeats itself over and over. Question #14 When are \frac{1}{n!} repeating decimal with single digit repetend Answer: A positive rational number has terminating decimal expansion iff it can be expressed as 𝑥 = 𝐴 10𝑘 for some 𝐴, 𝑘 ∈ 𝑁 A rational number has a length-1 periodic decimal expansion iff it's not in the form above, but 10𝑥 − 𝑥 = 9𝑥 = 𝐴 10𝑘 . Then, if 𝑥 = 1 𝑚 we require that 𝑚 = 2𝑎 5𝑏 3𝑐 with 𝑎 ≥ 0, 𝑏 ≥ 0 and 𝑐 ≥ 1 That is, it must be a multiple of 3, and it can be a multiple of 2 and 5, with no other ' factor. Then, if 𝑚 = 𝑛! , we must have 𝑛 ∈ {3,4,5,6} Question #15 What is the sum of number of digits of the numbers 22001 and 52001 Answer: It looks like you don't need logarithms or any calculator to solve this problem. Let's start. First, observe that the following inequalities hold: 10𝑚 < ⏟2001 2 < 10𝑚+1 𝑚+1~digits 10𝑛 < 5 ⏟2001 < 10𝑛+1 𝑛+1~digits You get, 10𝑚+𝑛 < 102001 < 10𝑚+𝑛+2 2001=m+n+1 m+n=2000 Finally, the sum of digits of 22001 and 52001 i equal: m+1+n+1=m+n+2 =2000+2 =2002 Question #16 What is the mistake in balancing the equation in a demonstration that 0.999 ⋯ = 1 ? Answer: There is no mistake in these proofs. Both of them are completely valid and proof that 1 = 0.9999 ⋯. This statement is, indeed, true. This can be sometimes confusing to people and it's completely normal that it feels weird at first. If you think about it in the following way, it might make sense: What would 1 − 0.9999 ⋯ be? The difference between these two numbers is smaller then any rational number. The only number that is smaller than any other is 0. Thus: 1 − 0.9999 ⋯ = 0 , which implies 1 = 0.9999 ⋯. Question #17 What is the last non-zero digit of ((⋯ (((1!)! + 2!)! + 3!) ! + ⋯ ) ! + 1992!) !? Answer: Let 𝑎𝑛 denote the last nonzero digit of n!; we need only consider 𝑛 ≥ 5 Prove 𝑎𝑛 ∈ 𝑆 ≔ {2, 4, 6, 8} Prove 𝑎10𝑘+5 = 𝑎10𝑘+14 Determine the order of the permutation 𝜎 of S for which 𝑎10𝑘+15 = 𝜎(𝑎10𝑘+14 ) Find a fairly small upper bound p on the period of 𝑎𝑛 Convince yourself 𝑎𝑝 , , ⋯ , 𝑎2𝑝−1 is a period. Explain why the original problem's answer is 𝑎𝑝 then compute it. Question #18 Convergence of (𝑏𝑛 ) if 𝑏 = 𝑏0 . 𝑏1 𝑏2 ⋯ 𝑏𝑛 ⋯ Answer: You're absolutely right that the eventually you need 𝑏𝑛 = 𝑏𝑛+1 = ⋯ since each 𝑏𝑛 can only be selected from 0, 1, ⋯ , 9 So either you're repeating 0's or repeating one of 1 through 9's. Lets assume for simplicity that 𝑏0 = 0 If you're repeating 0's, then 𝑏 = 𝑁 for some 𝑁 < 10𝑛 and n is the last nonzero digit. 10𝑛 If you're repeating k's for 𝑘 ∈ { 1 ⋯ 9} , then 𝑏= 𝑁 10𝑛 + 𝑘 9×10𝑛 where n is the last nonrepeating digit. For example, 0.58777777 … = 58 100 + 7 900 Note that this does not guarantee that 𝑁 10𝑛 or 𝑘 9×10𝑛 can't be simplified further. So, you could write (𝑏𝑛 ) converges if and only if there exists 𝑛 ∈ 𝑁, 𝑁 ∈ 𝑁𝟘 satisfying 𝑁 < 10𝑛 , and k ∈ {0, ⋯ 9} such that 𝑏 − 𝑏0 = 𝑁 10𝑛 + 𝑘 9×10𝑛 Question #19 Determine, with proof, the minimum value of c such that 𝑐 𝑛 + 2014 has all digits less than 5, where 𝑐, 𝑛 ∈ 𝑁 possible values of n. Note that 𝑁 does not include 0. In other words, c, n shall not assume the value of 0 or anything lesser than 1 in general. Answer: It is easy to verify that 10 works (assuming that 𝑁 is the positive integers - nothing works if n can be 0). Now 1 to 5 don't work because taking n=1 we have a value between 2015 and 2019 - the units digit is too big. Also, 6 to 9 don't work, because taking n=2 we have a value between 2050 and 2095 - the tens digit is too big. Question #20 Does parity of f(a) and f(a+1) are same whenever a is even? Question related to sum of digits. Answer: Assume a is even. That means we already know the parity with i=2j+1 odd: 𝐷(𝑎, 2𝑗 + 1) = 2𝑘𝑗 (even) and 𝑗 = 1, ⋯ 𝑎 2 −1 Now take a look for i=2j even: We don't know the concrete value of 𝐷(𝑎, 2𝑗), but we can draw a connection between D(a,2j) and D(a+1,2j) . As a is even the last digit in any even-numbered system must be even as well (otherwise we will get a contradiction mod 2)! This means the last digit can never be i-1 so if we add 1 we will only increase the last digit to a value ≤ 𝑖 − 1 and no overflow to the next digit is generated. So all digits besides the last one stay untouched. That means we get 𝐷(𝑎 + 1,2𝑗) = 𝐷(𝑎, 2𝑗) + 1 Summing all information up we can conclude: 𝑎 2 𝑎 −1 2 𝑓(𝑎) = ∑ 𝐷(𝑎, 2𝑗) + ∑ 𝐷(𝑎, 2𝑗 + 1) 𝑗=1 𝑗=1 𝑎 𝑎 −1 2 2 𝑓(𝑎 + 1) = ∑𝑗=1 𝐷(𝑎 + 1,2𝑗) + ∑𝑗=1 𝐷(𝑎 + 1,2𝑗 + 1) + 𝐷(𝑎 + 1, 𝑎 + 1) 𝑎 𝑎 𝑎 −1 2 2 = ∑𝑗=1 𝐷(𝑎, 2𝑗) + ⋅ 1 + ∑𝑗=1 𝐷(𝑎 + 1,2𝑗 + 1) + 1 2 Considering everything mod 2 we can use that 𝐷(𝑎 + 1,2𝑗 + 1) ≡ 𝐷(𝑎, 2𝑗 + 1) + 1 𝑎 𝑎 𝑎 −1 𝑎 2 2 𝑓(𝑎 + 1) ≡ ∑𝑗=1 𝐷(𝑎, 2𝑗) + ⋅ 1 + ∑𝑗=1 𝐷(𝑎, 2𝑗 + 1) + ( − 1) ⋅ 1 + 1 2 𝑎 𝑎 2 2 2 ≡ 𝑓(𝑎) + + − 1 + 1 = 𝑓(𝑎) + 𝑎 ≡ 𝑓(𝑎) 𝑚𝑜𝑑 2 Here we have it that 𝑓(𝑎) ≡ 𝑓(𝑎 + 1) 𝑚𝑜𝑑 2 if a is even.