### 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.

(c) Then tell them to subtract the two numbers (they can use a calculator if they wish).

(d) Ask them to circle any digit in the result except zero.

2. NO ELEPHANTS IN DEMARK TRICK.

This trick starts the same as the one above ... variations a, b and c can also be used!

(c) Ask them to subtract the two numbers (they can use a calculator if they wish).

(e) Tell them to add 4 to this number and then add the two digits together again.

(f) Ask them to use this number to get a letter of the alphabet such that A=1, B=2 etc.

3. THE ULTIMATE MISSING NUMBER TRICK.

(e) Ask them to circle any digit in the number selected except zero.

This trick works for the same reason that the first 'missing number trick' works - ‘digital roots’.

4. MENTAL MATHS AND NOT-SO-MENTAL MATHS.

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.

5. MENTAL AS ANYTHING (MORE MENTAL MATHS).

Sorry, I'll be posting a full explaination (with proofs) later.

6. DEVISING YOUR OWN DIVISIBILITY TESTS.

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.

7. FUN AND GAMES ON A CALCULATOR.

**(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!

Great tip! I just came across this blog while searching for what other CAT relevant sites had to say.

ReplyDeleteSharing my experience, i found great value in my personalized mentoring program for the CAT, which has helped me immensely to crack my CAT.. its helped me on 3 diff things - a. time b. accuracy and c. confidence.. which helped me perform well in my CAT..

Cheers,

Wow..really good post. I appreciate your work. Thank you for sharing.

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