Index: Xⁿ starts with n | | DJS

Index: Xⁿ starts with n

Often said in my lessons are that 2^6 is 64 and 2^10 is ten twenty four, 1024, and that these are the only 2ⁿ that start with n. [Prove!] This page explores the relationship between integers to the power of another integer, looking especially at those results which, like 2⁶ = 64, begin with the same digits as the index used. 

For example, 42⁵⁹ is 5.9x10⁹⁵ or 42⁵⁹ =  59x10⁹⁴, in a non-standard format.

Sometimes we will need general terms to describe this computation; I choose to refer, should I need to and using the example 42⁵⁹, such that  42 is the base and 59 is the index.

1. Can you point straightaway to some powers of 20 that will fit this property of 20ⁿ starting with n? Demonstrate that you're right and make a statement about other cases where the base number is a multiple of ten.

2. Thinking of easy squares —those you know well— spot the square that starts with a two. Write this down and also the similar result for ten times as much, which demonstrates the property you wrote down in Q1.

3. Using your calculator, show that there is a run of (two-digit) squares as the base number rises by one that also start with the digit two.

4. 2^10 is given at the top and is generally assumed knowledge (by me in my classes), so you might easily recognise that 8⁴ (= 2¹², = 4x1024) starts with a 4. More than this, you should now recognise that 80⁴ starts with a 4 and, quite possibly, 81⁴ does. Show that you can extend this towards 85 and show that 84 works but 85 does not.

5. Quite soon your calculator complains. By writing the values down in standard form, show that these calculations fit the requirement:

   31⁸,    58⁵⁶,    87⁵⁴,   99⁵⁶.

For some calculators, those last two will prove too hard. Explain why this is, and use what you learned in Q1 and Q2 to avoid the problem and still find the answer. Whatever solution you found, show it. One is enough.

6. Now you see a way to recognise whether you've found combinations that work, find which of these fit the property of having the first few digits (or non-zero digits if you're using numbers less than unity) matching the index:    86⁶⁴,  88⁵⁰,  470³²,  850⁴,  840¹²,  81⁸⁸,  82¹¹,  99855⁸⁸

The result you discovered in Q4 suggests that if a single digit number like 8 raised to a power like 4 fits, then it is likely that 8.1 and 8.2 will fit until some limit is reached.  

7. Find the simple 0.7ⁿ such the first non-zero digit is n and find the biggest number you can, quite close to 0.7, that still starts with that n.

8. 0.55⁵=0.050, so 0.551⁵ begins 0.5. The first number bigger than 0.55, X, that fails to fit the pattern here (numbers close to 0.55 that fit the property) is such that X⁵=0.06. [Obviously; if it doesn't start with 5 and it is a bit bigger, it must start with 6.] Find this X and consequently show the biggest integer close to 550 such that its fifth power will begin with the digit 5.      Does 549 work?

Around about now you begin to see that using numbers between zero and one is going to be an awful lot easier. So we might change 'fit the pattern' to 'the truncated decimal matches the index' and call our effect something like a 'truncated index match'. In class, we might call that a 'tim', especially if there was a Tim present.

9. The fifth power produces more numbers of this sort. 0.9⁵= 0.59049. By looking at the fifth root of 0.5 and of 0.6, find the range of integers close to 900 (above and below) whose fifth power starts with a 5.

Looking for more numbers where X⁵ begins with a 5 but not a 6, Q8&Q9 suggest that looking at the fifth root of 0.5, 0.05, 0.005 etc and 0.6, 0.06, 0.006, etc will reveal a range of three digit numbers that fit the pattern.

10. By writing the fifth root of 5x10⁻ⁿ and 6x10⁻ⁿ for various n, find the five groups of numbers below 10000 which will fit the pattern.

You now have a techniques for discovering a lot of numbers that fit this pattern. I decided to be interested in the fifteenth power, which produced a table like this on Excel.

The bottom row, which I'll call row 15, suggests to me that, of numbers below 1000, 103¹⁵=0.15.....

11. Use the table given ( or make your own) to write down which integers¹⁵ below 1000 will begin 0.15.

Here are the formulas I put into EXCEL, where I could have Named cell R4 as 'index', in which case it would look a lot tidier, like this:-

I've decided to be interested in the index being 37, as you can see.

12. Produce your own spreadsheet (this is now well into GCSE-rated project-work) and investigate. 

I say the digit integers such that X³⁷ begins 37... when truncated (i.e. fitting the 'Xⁿ starts with an n' pattern we've been investigating) include 63 and 86, but also 338, 383 and nine more three digit integers. 

The same spreadsheet showed that 0.9856^137 =0.137(0866), 0.9692^137 = 0.0137(057), 0.9063^137= 137.94x10⁻⁴  and so on. 766^137 works, but I think it is the only three digit solution.

I found it quite difficult to persuade my spreadsheet to calculate or show which integer solutions would work without it being me doing the decision-making. Maybe I was being thick, but at the time I found this hard.

Few students will complete all twelve questions. This might make up a week of lessons for some classes, a independent piece of coursework, or just a weekend homework.

This page as a whole could be seen as chasing numbers around on a calculator. It could also be seen as having a little problem that deserves just enough attention to cause a class collectively or individually to hunt for answers. It may serve to educate a class into the use of the xʸ button and its inverse, ˣ√y. Not least, it shows that while there may be 'answers' there is also a region of grey not-quite answers in the immediate neighbourhood. The more maths one does the less the subject is black and white and it becomes seen as more filled with many shades of grey. I would suggest that a whole-class exercise gives the feature to be pursued its own name (Tim, perhaps)— as I've written before, possession is nine-tenths of the learning.

If we did some maths first (rather than play with arithmetic) then all the exploration turns into preparation for understanding what comes next - the maths. For an index of 37,

we want   0.37≤X³⁷<0.38,   so    ³⁷√0.37≤X< ³⁷√0.38, 

which in turn means 0.973486≤X<0.974188 so that the integer 'solutions' are 974, 9735-41, 97349-97418 and so on. That approach (eventually) produced this next spreadsheet, where the index is 5 and the 3 indicates that solutions will be recorded from 1 to 999. The bold 5 is cell AC1 and is now called Nind, so the leftmost column is AB. I have a figure 3 in cell AE1 which is the number of digits (I thought of it as 'size') I want to have shown before a decimal point, as the next example shows. Beware filling downwards, for I've used pairs of lines (columns AC and AF) to achieve my objectives.

That same spreadsheet now shows all of the results for X⁵ = 5....., where X<1000. The top two 871 and 902 (on the right in Col AF) indicate that all the integers from 871 to 902 inclusive succeed in beginning with a five. While that 871-902 range implies 90 and 9 will also fit (because 900 does), filling the table downwards produces these answers too.

I swap the top left 5 to a 37 and instantly have the results for that different problem in descending order:  974, 915, 860, 808, 759, 630, 592, 491, 383, 338 and presumably 86 and 63. And no others. 1036 appeared in my table and works. I subsequently found some larger numbers where it is more difficult to show the valid integer range, such as where X³⁷ = 3.7x10⁵, for which 1415 works

I switch the index to 137 and only 766 shows. I change the size setting from 3 to 4 and get this quite long list:-   9856, 9692, 9062, 8616, 8056, 7660, 7162, 6925, 6475, 6367, 6054, 5953, 5566, 5473, 5382, 5204, 5032, 4948, 4399, 3977, 2560, 2402, 2362, 1805, 1166 and of course 766. 

DJS 20210531

1. Surely 20^6 = (2x10)⁶ = 2⁶x10⁶.  Similarly 20^6 = 1024x10⁶ = 10.24x10¹¹

5.  87⁵⁴ = 8.7⁵⁴x 10⁵⁴ = 5.4x10⁵⁰x10⁵⁴ = 5.4x10¹⁰⁴. Extending this 0.87⁵⁴ = 0.00054(2050) =5.4x10⁻⁴ so 87⁵⁴ = 5.4x10⁻⁴x10¹⁰⁸ = 5.4x10¹⁰⁴. Some will go to a bigger calculator, such as within Excel. That gets the answer but doesn't understand why the calculator won't do it. 

6. 8.4⁴ = 4978 so 84 and 840 both fit the property but 8.5 = 5220, so doesn't. Therefore 850⁴ doesn't either. You might prefer to work with 0.84⁴=0.4(9787) and 0.85⁴=0.5(2200625).

7. 0.7³ = 0.343, 0.73³=0.389, 0.736806300 is just too big. Think about how to persuade the calculator to tell you that. This isn't about chasing answers, it is about understanding how numbers fit together.

8. ⁵√0.06 = 0.569679052 so 569 works but not 570. Also ⁵√0.05 = 0.54928 so 55 and 56 both work, making the whole range here the integers from 550 to 569.

9.  ⁵√0.6 = 0.90288 so 901 and 902 will work but 903 will not. ⁵√0.5 = 0.87055 so 870 won't work but 871 will, and many bigger integers. So the range of three digit numbers here that work is 871 to 902. We could write this as 871 ≤ X ≤ 902.

10.  ⁵√0.6 = 0.90288  and ⁵√0.5 = 0.87055 so 871 to 902 from Q9

  ⁵√0.06 = 0.5696  and   ⁵√0.05 = 0.5492.  So 550 to 569 (not 549, not 570) from Q8

  ⁵√0.006 = 0.3594,        ⁵√0.005 = 0.34657. So 35⁵ works, and 3466⁵ to 3593⁵ will work

  ⁵√0.0006 = 0.22679,    ⁵√0.0005 = 0.21867 ; 22⁵ works and 2187 to 2267 will work

  ⁵√0.00006 = 0.143096; ⁵√0.00005 = 0.13797; 14⁵ works and 1380 to 1430 will work.

       Inserting yet another zero causes the pattern to repeat. Some will say "Of course it does" and others will need to hunt for why this is true. I suggest you write some of the numbers in standard form.

11. Working from the bottom row upwards gives the values in number order; 103, 120, 140, 163, 190, 222, 259, 301-2, 351-2, 410, 477-8, 556-8, 649-51, 756-9, 882-4 and, removing zeroes, 12,14 and 19 also work..

In November 2021 I threw this at a Y13 already known to be a good mathematician (and Oxbridge aspirant) and he found this hard. But then he had only had 20 minutes of horrible, off-the-wall me and not the backgrouind that any PMC Y8 top set student would have become entirely familiar with. So perhaps problems like this are harder than I imagine them to be. Alternatively maybe a whole classful of bright Y8 are actually sharing enough working brain cells (called teamwork if I'm polite) to produce some relatively brilliant results?

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