Hey come back with maths
long time not published any thing on this blog. But i am back, can't leave blogger, with more important tricks. Now from here we will have some real time practice on all subjects.
In this post we will see some maths for CAT exam. These tricks are extremely useful so learn it by heart.
(f) To find the missing digit, simply add up all the digits that they give you - repeatedly - until a single digit 1 to 9 is obtained. If this value is 9, then the number they circled was 9. Otherwise, add a value to the number until it is 9. This value is the number they circled! e.g. If the digits are 9, 8, 8 and 4, then 9+8+8+4=29, 2+9=11, 1+1=2. 9-2=7. Therefore the number circled was 7.
(a) Ask them to write the largest number possible (10 or maybe 20 digits) and repeat the trick. Remember, if they are using a calculator, the maximum number of digits that can be entered on the calculator may be 8, 10 or 12 digits.
(b) Instead of rearranging the digits, ask them to add up the digits and subtract this total from the number. This may also be done as an extra step! Or, for a real interesting variation, ask them to add up only the odd digits (units, hundreds ...) then subtract this value from the number. Then add up the even digits (tens, thousands ...) and add this to that result - then add it again. Finally, multiply the result by three!
(c) As an extra step (before or after the subtraction), ask them to subtract a number (e.g. 1, 2 ,3) from one digit and add it to another digit - even having a carry or borrow is OK! They could also subtract a number from one digit and place this value as an extra digit anywhere in number (on the left or right of the other digits is best). Also as an extra step (after the subtraction), you can ask them to add the number 3456 for example. Once you discover the 'secret' of this trick you'll see that many other numbers can be 'safely' added or subtracted and the trick still works! Finally, as an interesting variation, ask them to remove any digit from the number (cross it off) multiply this digit by 8 and then subtract this value from any digit(s) in the number. You can even ask them to split the value into two digits and subtract these separately anywhere in the their number!
(f) Instead of adding the digits mentally, enter them on a calculator and then divide by 9. If there is no decimal part, the number circled was nine (or zero if the variation above is used). Otherwise, the digit to the right of the decimal point is subtracted from nine to get the digit circled!
(g) This trick also works with other number systems. e.g. octal (base 8) or hexadecimal (base 16). Only one small variation is needed to determine the missing number. What is it? Most people can’t add or subtract in other number systems mentally, so you might like to try this with a calculator such as the Casio fx-100 which does binary, octal and hexadecimal calculations. You should note that binary (base 2) is useless for this trick...why?
This is a truly great trick that can be used to amaze your friends and relatives and it works very simply. How? Well, I’m going to give a full explanation (with proof) later. For now, this trick works because of what is called ‘digital roots’. There are other interesting things you can do with ‘digital roots’ which I’ll also explain over the coming weeks and months.
This trick works for the same reason that the one above works - ‘digital roots’. The country and animal part is just a bit of window dressing ... you will see immediately how this part of the trick works after you do it a couple of times!
This trick is a bit more difficult then number 1 above, but once you discover 'the secret' and once you have practised this trick a few times (perhaps using pen and paper to begin with) you'll find it quite easy!
(b) Ask them to add up all the digits of the number and tell you the result. Now, you don't need to remember this result, just continue adding the digits of the number that they tell you until you get a single digit 1 to 9 only. e.g. for 29, 2+9=11, 1+1=2, just remember 2! Call this number N.
(c) Then tell them (trying to seem like you are making these numbers up as you go) to add 6543 and write down the result. Then subtract 567 and write down this result. You can also continue and ask them to add or subtract other numbers like 432. Once you discover the 'secret' of this trick you'll see that many other numbers can be 'safely' added or subtracted and the trick still works!
(d) Now, as an option, you can let them choose which of the above results they wish to use. But whatever, ask them to add up the digits of the number and satisfy themselves that it is not the same as before!
(g) To find the missing digit, simply add up all the digits that they give you - repeatedly - until a single digit 1 to 9 is obtained. If this value is the same as the number N (see b. above), then the number they circled was 9. Otherwise, subtract this value from N. If this result is positive this is the number they circled! If negative, add 9 to the value and this is the number circled! e.g. If the digits are 9, 8, 8, and 4 then 9+8+8+4=29, 2+9=11, 1+1=2, If N=5, then 5-2=3 Therefore the number circled was 3. If N=1, then 1-2=-1, -1+9=8, the number circled was 8.
Here I'm going to explore some of the ways that 'digital roots' can be used in mental arithmetic. Perhaps you could use it at school (as a teacher or student) or at home to amuse yourself or the kids. Whatever, I find these techniques fascinating and I only wish I knew about them when I was at school! I should explain, first up, that what we are using here is the 'additive digital root' or 'digit sum'. Which is, of course, where you add the digits. There is also the 'multiplicative digital root' where you multiply the digits - however, I would suggest that if the same 'rules' were to apply - ALL multiplicative digital roots would be ZERO! In any case, just like 'perfect numbers', 'happy numbers' and 'vampire numbers' and other weird and wonderful things that mathematicians come up with, I'm yet to find ANY sort of use for multiplicative digital roots! If you'd like to find out more about this and other 'absolutely useless' numbers just go to - mathworld.wolfram.com. I am also going to explore other forms of 'digital roots'. One, which I call the 'true digital root' or 9's digital root and others I call the 3's digital root and the 'alternating digital root' or 11's digital root (I don't know if anyone else has come up with this before, but a quick search of the internet suggests not).
(a) The nine times table... The first thing you'll discover about digital roots is that if it equals 9 then the number is divisible by 9. Let's see: 9, 1+8=9, 2+7=9, 3+6=9 ... But what about 9+9=18, don't forget we need to keep going until we have only a single digit. So 9+9=18, 1+8=9 OK? Well, I don't know about you, but the way I did my nine times table was to multiply by ten and then subtract. i.e. 9x7 = 70-7 = 63. But once you know that the two digits of the number MUST add to 9 it becomes MUCH easier! Try it and see. In fact, using this technique, your 9 times table becomes almost as easy as the 11 times table!
Here's another fact that will help you with your 9 times table. Any number N multiplied by nine, if you subtract the units digit of N from 10 this will equal the units digit of the result. e.g. 8 x 9 = 72, 10-8=2, note the unit digit of the result is 2. Try 12 x 9 = 108, 10-2=8, units digit of result is 8! This fact always works except when the units digit of N is 0, then the result units digit will also be 0.
We can use this fact to help us multiply all numbers from 11 to 99 by 9. Of course you could simply use long multiplication or multiply by 10 and subtract but you might find this next technique easier. If multiplying N x 9, take the tens digit of N and add 1. Subtract this value from N, this is the first part of the answer. The final units digit is worked out as above. OK, let's try 12 x 9, 1+1=2, 12-2=10, 10-2=8, result is 108. Get it? Another one, 63 x 9, 6+1=7, 63-7=56, 10-3=7, answer 567. How about 99 x 9 = 891 - did you get this? Of course there's the special case of 0. It's best to remove the 0, multiply by 9 and then add the 0 to the result. i.e. for 70 x 9, 7x9=63, answer 630.
If you know your nine times table up to 5x9=45 but are having trouble with 6x9 up to 9x9 then here's a little curiosity that might help! Just subtract the number from eleven, multiply that by 9 and swap the digits around (e.g. switch the tens and units digits). Let's see? 8x9, 11-8=3, 3x9=27, therefore 8x9=72 easy! Again, 7x9, 11-7=4, 4x9=36 then 7x9=63 right?
Finally, if you've forgotten your 9 times table just enter on a calculator (minimum 9 digit calculator is required) 102030405 and multiply by 9, the answer is 918273645. Get it? 9, 18, 27, 36, 45. Then reverse the digits. 54, 63, 72, 81 and 90 (i.e. imagine there's a leading zero). For other interesting tricks on a calculator see: fun and games on a calculator below.
(b) Divisibility by nine and remainders... You may wonder why divisibility tests are important. If you wish to be quick and efficient in maths then recognising divisibility is important when simplifying fractions, simplifying or solving algebraic equations and for pattern recognition...plus more! I think most people know that any number divisible by 2 will end in 0,2,4,6,8 (even numbers), divisible by 5 will end in 0 or 5 and divisible by 10 will end in 0. And, with a little bit of thought, you'll probably be able to work out the remainder of any number divided by 2, 5 or 10. But what about other numbers? Digital roots, once again, allow us to quickly work out, not only, if a number is divisible by 9 but also the remainder - if necessary. Let's try the number 3644. 3+6+4+4=17, 1+7=8. This number is NOT exactly divisible by 9, it would have a remainder of 8. In fact I can also tell you that 3645 is the next number divisible by 9 because 8+1=9. And that 3636 is the prior number divisible by 9 (just subtract 8). You should also note that any number divisible by 9 has a digital root of 9 - the remainder is 0 not 9 in this case. Now, for the 'true digital root' or 9's digital root. This method makes it quicker and easier to work out the remainder or divisibility of a number by 9, particularly if you have a pen and paper handy and can write down the number and cross off digits. The rule is simple - 'any digit which is nine is crossed off and replaced by zero'. You can even extend this further by saying any combination of digits that sum to 9, 18, 27 etc. can also be crossed off! I can now write a number like 1234567892 on paper and within a few seconds tell you that this number has a remainder of 2 if divided by 9. How? First cross off the 9, then 1+8, 2+7, 3+6, 4+5 and all you have left is 2. Or, if you prefer, as soon as the sum equals or exceeds 9 - subtract 9 (or add the two digits) and continue. Remember, with this method, you can never be left with 9 because it is simply crossed off and replaced with zero - automatically giving you the correct remainder!
In fact, you should never have to add ANY sums bigger than 8+8=16, because 16-9 or 1+6=7. You can simply continue with 7 added to the next digit etc. For example with 567862. 5+6=11, 1+1=2, 2+7=9 (becomes 0), 8+6=14, 1+4=5, 5+2=7. Or maybe it's easier with, 5+6+7=18 (becomes 0), 8+6+2=16, 1+6=7. You decide!
Finally, after first learning about digital roots, I wondered how useful they'd be for solving problems. Then I came across a question on an ACER scholarship test, which read: 'In Russia a calendar was introduced in 1923 in which all years had 365 days, except those which, when divided by 9, had either 2 or 6 as a remainder. These were given an extra day (leap years). Which of the following was a leap year in Russia? A. 1929 B. 1934 C. 1936 D. 1937.' Well, I could almost see the answer 'pop out' at me! D. By-the-way, as far as I'm aware, this question is not entirely factual. The Russians did devise a system, like this, however to work out which century years (1900, 2000, 2100 etc.) would be a leap year which was more accurate than the Gregorian calendar we use today! e.g. the Russian leap years would be in 2000, 2400, 2900, 3300, 3800 ... By-the-way, if you'd like more information about ACER or ITSA scholarship tests - click here.
(c) Divisibility by three and remainders... I'm sure that anyone can recognise that 3 divides into 9 exactly. So, if we divide the digital root by 3 we will get the remainder. If that remainder is zero then the number is divisible by 3, otherwise we will get 1 or 2. Let's try the number 1416. 1+4+1+6=12. 1+2=3. 3/3=1 with no remainder! Again, there's a simplier way of working this out. Obtaining what I call the 3's digital root. In this case you cross out any digit that is 3, 6 or 9. And any combination of digits that sum to 3, 6, 9, 12 etc. Let's try the number 3485452612. Cross off 3, 6. Also 4+8, 5+4, 1+2. You are left with 5+2. 7/3 = 2 with 1 remainder. Simple. Or, if you prefer, cross off 3, 6 or 9. Replace both 4 and 7 with 1's and replace both 5 and 8 with 2's (the remainders if divided by 3). Then cross off any pairs of 1 and 2, then simply add up any 1's or 2's remaining - divide that by 3 and bingo!
Finally, out of interest, there is another (more complex) method of testing for divisibility by 3 - AND for finding the remainder! First add all the ODD positioned digits (let's call it A), then add all the EVEN positioned digits (B). If A-2B is divisible by 3 so is the original number! If it's the remainder you're after, the result must be between -9 and +9 (a single digit), otherwise do it again and again if necessary. Once you have a single digit, divide it by 3 to obtain a remainder. If it's negative, add 3 giving a value of positive 1 or 2. Let's try 82134682. A=2+6+3+2=13, B=8+4+1+8=21, A-2B=13-42= -29, -(9-4)= -5, -5/3 = -1 and -2 remaining, -2+3=1. The number has a remainder of 1 when divided by 3! Let's try another, 213468. A=1+4+8=13, B=2+3+6=11, A-2B= 13-22= -9. This number is exactly divisible by 3. One interesting observation about this method is that it implies that any two digit number where one digit is twice the other digit will ALWAYS be divisible by three. Let's see - 12, 21, 24, 42, 36, 63, 48 and 84 - all divisible by three! It also implies that any three digit number where the digits are the same will also be divisible by three. 111, 222, 333 ... and 999 are all divisible by three!
(d) Divisibility by eleven and remainders... To test whether a number is divisible by 11 or to get the remainder, we need to find, what I call, the 'alternating digital root' or the 11's digital root. Start by doing what we did with the second method for divisibility by 3. Add up all the ODD positioned digits (A), then add up all the EVEN positioned digits (B). Subtract A-B, stop if the result is between -9 and +10 (you can stop earlier if any result is exactly divisible by 11 or you can determine the remainder) , otherwise do it again and again if necessary. Finally, if the result is zero or positive that is the remainder when divided by 11. If the result is negative, add 11 and that becomes the remainder (from this you will see that if the result is -1, adding 11 will give a remainder of 10. If the remainder is zero then the original number is divisible by 11. Please note, if the remainder is required the SIGN of each result is important and must flow through the calculations correctly! e.g. Let's try 567862. A=2+8+6=16. B=6+7+5=18. A-B=16-18=-2. The remainder, if divided by 11, is -2+11=9. Again, methods can be used to simplify the procedure. First, you can cross off any SAME two digits in odd and even positions. You can go further and cross off any combination of odd or even positioned digits which sum to 11, 22, 33 etc. Or, if you prefer, as soon as the sum equals or exceeds 11 - subtract 11 and continue.
Just as with the test above for divisibility by 3, you should be able to make some interesting observations about numbers divisible by 11. With the A-B formula, any two digit number where the digits are the same will be divisible by 11. e.g. 11, 22, 33... (pretty obvious). But how about numbers like, 123321 or 1234554321 or 123456789987654321? Test them and see.
Note, if you are only interested in testing for divisibility by 11 and NOT the remainder then it doesn't matter whether you do A-B or B-A ... you can just ignore the sign anyway as long as the result is zero! If however, you wish to find the remainder but have forgotten whether it is evens from odds or odds from evens just think of the number 12. when divided by 11 the remainder is 1...which is 2-1 (i.e. evens from odds). If you do it the opposite way 1-2= -1, the remainder is -1+11= 10 (obviously wrong). By-the-way, I have called this the 'alternating digital root' because you are effectively adding the unit's digit and then subtracting the ten's digit and adding the hundred's digit and so on. Get it?
How does this all work?
Sorry, I'll be posting a full explaination (with proofs) later.
(a) The six, seven and eight times table... I think most people have difficulty learning (and remembering) their 6, 7 and 8 times tables. Note, the 'trick' for the 9 times table is discussed above. Since we really only need to deal with products involving these three numbers, all you need to remember is 6x6, 6x7, 6x8, 7x7, 7x8 and 8x8 ... that's all! Here's a trick to help you remember. Let's take 7x8. Subtract 7 from 10 = 3 and 8 from 10 = 2. Now, 7-2 or 8-3 = 5 (that's your tens digit) and 2x3=6 (that's your units digit) i.e. 56. Let's try 6x8. 10-6 = 4, 10-8 = 2, 8-4 or 6-2 = 4, 4x2=8, answer 48. Doing 6x7 is a little bit more difficult. 10-6=4, 10-7=3, 6-3 or 7-4 = 3 and 4x3=12. Therefore 30+12=42. Squaring 6, 7 or 8 (6x6, 7x7 or 8x8) is even easiler. With 6x6, 10-6=4, 6-4=2, 4x4=16, 20+16=36. With 7x7, 10-7=3, 7-3=4 and 3x3=9, answer 49. With 8x8, 10-8=2, 8-2=6, 2x2=4, answer 64. Alternatively, instead of doing 10-8=2 and 8-2=6, use 8+8 or 8x2 = 16 and take the units digit (i.e. subtract 10).
P.S. I'm not sure how 'practical' the above method really is, it does however provide an interesting introduction to (b) below. A more practical method might be to get the 5 times value and add. e.g. 6x6 = 5x6+6 = 36. 7x8 = 8x5+16 = 56. etc.
(b) Multiplying numbers between 11 and 19... Now, to do this in your head, we are going to use a similar technique to that in (a) above. As long as you know your times tables and can add some simple sums you'll be able to do this - with practice! First let's try 12x13. 12-10=2 and 13-10=3 (i.e. just take the units digits). Either add 13+2 or 12+3 = 15 (this is the tens digit = 150). Multiply 2x3 = 6 (this is the units). Add 150+6 = 156 is the answer. Let's try 17x18. Add 17+8 or 18+7 = 25. Multiply 8x7 = 56. Add 250+56 = 306 is the answer. How about squaring numbers between 13 and 19...easy! Let's do 16 squared. 16+6=22, 6x6=36, then 220+36 = 256 is the answer!
P.S. Once you have mastered doing this in you head, you will then be able to multiply any numbers between 0 and 20 in your head. But what about single digit (2..9) multiplied by double digit numbers? Easy. You can simply use long multiplication! Here's how. Try 8x19 = 8x9 + 80 = 152. Try 6x14 = 6x4 + 60 = 84.
(c) Multiplying any two digit number by 11... Any two digit number (10 to 99) can be multiplied by 11. Simply by adding the digits together and placing this in the tens column, whilst the tens digit goes in the hundred's column and the units digits remains. e.g. Let's try 45x11 = 4(4+5)5 = 495. 32x11 = 3(3+2)2 = 352. 65x11= 6(6+5)5 = 715. 99x11 = 9(9+9)9 = 1089. Pretty simple eh?
(d) Dividing by 9... Any two digit number (10 to 99) can be divided by 9, the result is simply the tens digit plus the sum of the digits as the remainder. If the remainder is 9 or greater divide it by 9 and add the result to the previous one. e.g. divide 71 by 9. result is 7 with 7+1=8 remainder = 7r8 = 7.8888.... 84 divided by 9 = 8 with 8+4=12 remainder, since 9 divides into 12 once with 3 (1+2) remaining the answer is 9r3 = 9.3333.... 97 divided by 9 = 9r16 = 10r7. For 3 digit numbers (100...999) the result is the hundred's digit added to the tens with the sum of digits as the remainder. Again if the remainder is 9 or greater divide by 9 again. e.g. 123 divided by 9 = 12+1=13 with remainder 1+2+3=6, 13r6 = 13.6666.... 517 divided by 9 = 51+5=56 remainder 5+1+7=13 = 57r4 = 57.4444.... Larger numbers can be divided by 9 simply by added the higher digit to the next lower repeatedly! And, summing the digits provides the remainder...if the remainder is greater than 9 divide this by 9 again etc. e.g divide 1245 by 9 the result is 124+10=134, 134+3= 137. With remainder 7+5 = 12. 137r12 = 138r3 = 138.3333.... Of course, it get's harder when digit sums become greater than 9. e.g. 6782 divided by 9. 6+7=13, 13+8=21, 21+2=23. Answer: 600+130+21= 751 and 23 remaining = 753r5.
(e) Multiplying numbers between 91 and 109... Let's start with numbers between 101 and 109, this is even easier than multiplying numbers between 11 and 19! First let's try 102x103. 102-100=2 and 103-100=3 (i.e. just take the units digits). Either add 103+2 or 102+3 = 105 (this becomes the first 3 digits of the answer). Multiply 2x3 = 6 (this is the last 2 digits of the answer). e.g. 102x103=10506. Let's try 107x108. Add 107+8 or 108+7 = 115. Multiply 7x8 = 56. 11556 is the answer. How about squaring numbers between 101 and 109...easy! Let's do 106 squared. 106+6=112, 6x6=36. 11236 is the answer!
Multiplying numbers between 91 and 99 is almost as easy. Let's try 96x98. 100-96=4, 100-98=2, then either subtract 96-2 or 98-4 = 94 (this is the first 2 digits of the answer). Then 4x2=8 (is the last 2 digits of the answer). e.g. 98x96=9408. Squaring is easy. 91 squared is 91-9=82. 9x9=81, answer 8281.
The hardest thing to do is to multiply a number less than 100 (91..99) by a number greater than 100 (101..109). Let's try 92x104, 100-92=8, 104-100=4, 92+4 or 104-8 = 96. Then subtract 8x4=32 from 9600 = 9568 is the answer. It is easier with numbers of equal distance from 100. e.g. 95x105, 100-95=5, 105-100=5, 95+5 or 105-5 =100, subtract 10000-5x5= 9975. Easy?
(a) Combining divisibility tests... It should be pretty obvious that tests for other numbers can be obtained by combining divisibility tests. For example to test for divisibility by six. Simply test for divisibility by two and three (because 2x3=6). But what about divisibility by four? We know that the number must first be divisible by two (because 2x2=4). We could then divide the number by two and check that the result is even (last digit 0, 2, 4, 6 or 8). That's one way! However, we also know that 100 divided by 4 is 25. Therefore if the last two digits can be divided by four - so can the number! More specifically, if the tens digit times 2 plus the units digit is divisible by 4 so is the number. So now we can use this test and the test for three to test if a number is divisible by 12 (because 3x4=12). But what about divisibility by eight? Well, the number must be divisible by two (even) and the number must be divisible by four. We could divide the number by two and then test if the number is divisible by four (because 2x4=8), right? However, we also know that 1000 divided by 8 is 125. Therefore the last three digits must be divisible by eight! More specifically, if the hundreds digit times 4 plus the tens digit times 2 plus the units digit is divisible by 8 so is the number.
(b) Other divisibility tests... We can see by (a) above that most divisibility tests are now covered. The exception being prime numbers such as 7, 13, 17 ... So, here's the trick for divisibility by prime numbers (or multiples of). First find the first multiple of the number (let's call it N) that ends in 9 or 1 (e.g. 9, 11, 19, 21, 29...). Second, round up or down the number to the nearest multiple of ten (10, 20, 30...). Now your multiplying contant (let's call it K) becomes the tens digit (1, 2, 3...). If you rounded up it is positive (for add), if you rounded down it is negative (for subtract). Now take the number you wish to test for divisibility and multiply the units digit by K. This value is then added to the tens column or subtracted if K is negative. If the number that results is divisible by the number N then so is the original number! Of course, by logical extension, this process can be repeated until you recognize a number that is divisible by N.
Let's see if we can use this method to find a divisibility test for 7. First, multiply 7 until we have a number ending in 9 or 1. OK. 7, 14, 21. Now, we need to round this down to 20, so K = -2. Lets' test it. Try 3199. 319 - 2x9 = 301. 30 - 2 = 28. 28 is divisible by 7 so is 3199. Note, we could have gone further with 28. As 2-16 = -14. And 1-8 = -7. The factor K = -2 is also a test for divisibility by 21. The difference, being that we need to check if the result is divisible by 21 rather than 7! Let's try 9576. 957 - 2x6 = 945. 94 - 2x5 = 84. 8 - 2x4 = 0. The number 9576 is divisible by 21 (and 7 and 3 in fact). A couple of very interesting observations can be made. Firstly, all two digit numbers where the tens digit is twice the units digit (21, 42, 63 and 84) are all divisible by 21 (and by default 7 and 3). You should also notice the similarity of this test with the second test for divisibility by 3 above. In fact, this test can also be used to test for divisiblity by 3!
Alternatively, when testing for divisibility of 7, since 7x7 = 49, we can use K=+5. Let's try this with 294. 29 + 5x4 = 49. 49 is divisible by 7 and so is 294! Interestingly, we can't go any further with 49, since 4 + 5x9 = 49!
Note, despite what some books may tell you, tests for divisiblity by seven are not very 'practical'. I would recommend simply dividing the number by seven (you can discard the result if you're only interested in the remainder) and if the final remainder is zero then the number is divisible by seven! If you are using pen and paper then the number can be simplified first by crossing out 7's or any 2 digit multiple of 7 (make them 0), changing 8's to 1 and 9's to 2 (i.e. the remainders when divided by 7) and, since 7 divides into 1001, 10010, ... exactly, subtract one digit from another which are separated by 2 digits. Remember any leading or trailing zeros can be removed. e.g. Try 59633, cross off 63 = 59003, make 9 a 2 = 52003, subtract 3 from 2 = 49000, 49 is divisible by 7 so is 59633. Note, after obtaining 59003, the result could be obtained easier by taking the 3 from the 9 = 56000, 56 is divisible by 7! Also, since 1001 is a multiple of 7, K= -100. Which is the same as subtracting it from the 4th digit!
How about a divisibility test for 13? First 13, 26, 39. Round up to 40, therefore K = +4. Testing it, let's try 585. 58 + 4x5 = 78. 7 + 4x8 = 39. 39 is divisible by 13 and so is 585. What's real interesting about this method is that if testing for divisibility by 3 or 9, the value of K = +1. In other words, you repeatedly add the units digit to the tens column. If you continue to do this you end up with the digital root of the number! The value of K for other prime numbers is; K= -1 for 11. K= -5 for 17. K = +2 for 19. K= +7 for 23. Note, it's pretty easy to work out the values for K as required. i.e. There's no need to memorize them!
Finally, you may wish to find a negative value for K rather than use a positive value (the number reduces quicker and therefore the answer is found quicker). Simply subtract the number N from K. e.g. for N=3 and K=+1, the alternate value for K is 1-3 = -2. Try it and see if it works! To convert a negative K to a positive we can use N + K. e.g. for N=7 and K= -2, the alternate value of K is +5. For N=9, we can also use K= -8. For N=11, K= +10. etc. Note, with K= +10 for divisibility by 11, we can simply take the units digit and add it to the thousands column, if the resulting number is divisible by 11 so is the original number! Let's try 517, (5+7)1 = 121 we could stop here, but if we do it again (1+1)2 = 22. Clearly divisible by 11. Note, you can't go any further with this method (K= +10) once you have a 2 digit number. Of course, you could always continue with K= -1.
(b) Using divisibility tests to check for prime numbers... To test if a number is prime (divisible by itself and one only), you only need to test if it is divisible by other prime numbers up to the square root of the number you are testing. So, let's say we'd like to find all the prime numbers up to 100. We need to know all the prime numbers up to 10 (i.e. the square root of 100). They are 2, 3, 5 and 7. Now we know that: (a) All even numbers greater than two can be eliminated (they divide by two). (b) Starting from 9 we can eliminate every third odd number (e.g. 9, 15, 21 ...) as they are all divisible by three. Or, if you prefer the digital root of these numbers is 3, 6 or 9. (c) All other numbers ending in 5 can also be eliminated as they divide by five. (d) Finally, any number divisible by 7 can be removed. There is only three numbers here, 7x7= 49, 7x11= 77 and 7x13= 91. (Note, 11 and 13 are the next two prime numbers).
You can continue, if you wish, testing for prime numbers up to 169 (13x13) adding the test for divisiblity by 11. And, as long as you know that 169 is not a prime number, you can continue until 221 (13x17). You can go further using divisibility tests for 13, 17 and 19 etc. Finally, to generate prime numbers on your computer take a look at the QBASIC program on our puzzle page.
How does this all work?
Sorry, I'll be posting a full explaination (with proofs) later.
Here we are going to explore some of the things you can do on a calculator. Things that you may not have thought about. A simple add, subtract, multiply and divide calculator is needed. For the advanced exercises you will need a modern scientific calculator. I would suggest that if your calculator is more than a few years old you might wish to update to a new 2 line type (Casio FX-82 or FX-100, Texas TI-30 or Sharp EL-531). They are now commonly used in primary and junior secondary schools and cost as little as AUS$16 and are well worth it! You may also wish to use the calculator on your PC. In Windows, look under 'Accessories' - either an ordinary or scientific calculator can be selected.
(a) Digital roots on a calculator ... You might think that to work out the digital root of a number you would need to add each individual digit and then repeat the process until a single digit is obtained - BUT NO! All you need do is to enter the entire number on the calculator and divide it by nine. Then look at the digit to the right of the decimal point (d.p.) that represents the remainder of the number when divided by nine! Note, if there is nothing to the right of the d.p. then the number is exactly divisible by nine and the digital root is nine. Also ensure that at least two digits to the right of the d.p. are shown because the calculator may round up the number giving you an incorrect result. Finally, make sure that the number is NOT displayed in scientific notation!
To try this out, let's see if we can solve the problem in 4(b) above. Divide 1929 by 9 = 214.333.. The digital root or remainder when divided by 9 is 3. Divide 1934 by 9 = 214.888..9 Divide 1936 by 9 = 215.111.. Divide 1937 by 9 = 215.222.. The remainder or digital root is 2, therefore 1937 is a leap year - problem solved!
(b) Alternating digital roots on a calculator ... This is simply the remainder when a number is divided by 11. So, first divide your number by 11. Then look at the digit to the right of the decimal point (d.p.) and ADD ONE - that represents the remainder of the number when divided by eleven! Of course, if there is nothing to the right of the d.p. then the number is exactly divisible by 11 and the alternating digital root or remainder is zero.(c) Finding remainders on a calculator ... This is something that most calculators don't do well at all! Special case solutions are shown in (a) and (b) above, but what about generally finding remainders? On your computer's scientific calculator (see 'Accessories' and select 'View..Scientific') you will find a function called 'Mod' which is short for Modulus. Enter '11 Mod 9' and the answer is 2 etc. On a normal calculator you can do the division, then remove the integer part (i.e everything on the left of the d.p.) by subtracting it and then multiplying by the divisor. Unfortunately, this is all very clumsy but is made easier if you use a 2-line calculator such as the Casio FX-82. This calculator also allows you to display the result as a fraction (press a.b/c) in which case the numerator is the remainder (if, and only if) the denominator is the divisor! HINT, HINT, one might think that a smart calculator manufacture would provide the 'Mod' function given that it's commonly available on your computer!(c) Finding remainders on a calculator ... This is something that most calculators don't do well at all! Special case solutions are shown in (a) and (b) above, but what about generally finding remainders? On your computer's scientific calculator (see 'Accessories' and select 'View..Scientific') you will find a function called 'Mod' which is short for Modulus. Enter '11 Mod 9' and the answer is 2 etc. On a normal calculator you can do the division, then remove the integer part (i.e everything on the left of the d.p.) by subtracting it and then multiplying by the divisor. Unfortunately, this is all very clumsy but is made easier if you use a 2-line calculator such as the Casio FX-82. This calculator also allows you to display the result as a fraction (press a.b/c) in which case the numerator is the remainder (if, and only if) the denominator is the divisor! HINT, HINT, one might think that a smart calculator manufacture would provide the 'Mod' function given that it's commonly available on your computer!