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Math tricky 9

chatrack
Hai I have some cool trick with 9 !

step1. Immagine a number between 1 and 9 (single digit)

step2. multilpy that number with 9

step3. If the result contain ONLY 1 digit continue to step 4. If result have 2 digits in it ADD the two digit

step4. now add 1 to it

mtorregiani
It's a nice trick. But very easy to find out why's 10 the result.
_AVG_
Yes ... but it is a divisibility test of 9 ... all multiples of nine have digits that add up to nine. Hence, if you add the digits and then add 1 to them, it will always come up to be 10. Similarly, all multiples of 3 have digits that add up to 3.
friuser
it's only a trick if you didn't go to school.
SlideR.nl
 friuser wrote: it's only a trick if you didn't go to school.

That's right. It is just simple math. Write them all out and you see it's not a trick it's just simple math. But it is a nice found you did. It didn't come up to me, not that I was trying to think about something like that.
Bikerman
There are lots of nice little tricks you can do with numbers. Here's another simple one;
Want to multiply by 11?

Add the digits and stick the sum in the middle.

Hence:
23*11 = 2 (2+3=5) 3 = 253
41*11 = 4 (4+1=5) 1 = 451

Now - what happens when the digits sum to greater than 9? Try it.
AftershockVibe
Similarly, you can add up the indidivual digits of numbers to determine whether they are divisible by three.

For any number divisible by three, the digits which it is composed of will also sum to a number divisible by three.

e.g
21 ... (2+1) = 3

342834729325 ... (3+2+8+3+4+7+2+9+3+2+5) = 48 .. (4+8 ) = 12 ... (1+2) = 3

James007
Moved to Science basics.

I agree with friuser.
nilsmo
The reason why the remainders of the sum of the digits of a number x equals the number itself when you divide them by nine is algebraic:

x = a + 10*b + 100*c + 1000*d + ... = a + b*9+b + c*99+c + d*999+d + .... which divided by 9 gives a remainder of a + b + c + d + .... = sum of the digits
_AVG_
 Bikerman wrote: There are lots of nice little tricks you can do with numbers. Here's another simple one; Want to multiply by 11? Add the digits and stick the sum in the middle. Hence: 23*11 = 2 (2+3=5) 3 = 253 41*11 = 4 (4+1=5) 1 = 451 Now - what happens when the digits sum to greater than 9? Try it.

You can use the same rule ... here's an example.

79*11 = 7 (7+9) 9 but you cant write 7169. Hence you add 1 to 7 and put a 6 in the middle.
Bikerman
Correct - in other words you 'carry' the tens column to the left (just as you do in normal addition).
snowboardalliance
 friuser wrote: it's only a trick if you didn't go to school.

HAHAHAHA
Yeah, just agreeing that it's really basic, the rule that adding the digits of multiples of nine repeatedly will always be nine.
infinisa
Hello nilsmo,

Thanks for supplying this simple elegant proof:

 Quote: The reason why the remainders of the sum of the digits of a number x equals the number itself when you divide them by nine is algebraic: x = a + 10*b + 100*c + 1000*d + ... = a + b*9+b + c*99+c + d*999+d + .... which divided by 9 gives a remainder of a + b + c + d + .... = sum of the digits

Everyone else seems to think the matter is too trivial to deserve one. However, the fact that we all learnt it at school ("everyone knows...") shouldn't prevent us from asking "why?"

Some more questions on this theme:
1. Why does the rule also work for 3 (i.e. A number is divisible by 3 if and only if the sum of its digits is a multiple of 3)
2. Why does this "trick" only work for 3 & 9?
3. Find a similar trick for division by 11
4. What happens if you write numbers to base other than 10?

1. The rule works for 3 using a proof identical to the above, since b*9+c*99+d*999, being divisible by 9, is also divisible by 3
2. It's easy to see from the above proof that the key thing is that 9 must be a multiple of our number. This is true for 3 & 9 only.
3. A trick for divisibility by 11 is: if our number has digits abcde for example, take a-b+c-d+e (i.e. put in alternating - & + signs). The number is divisible by 11 if this new number is. Eg 1276 gives 1-2+7-6=0, so 1276 is divisible by 11 (1276=11x116).
4. The reason why 9 is special is because it is 10-1 (and 3 is a factor of 9). This is an "accident" due to our numbers being to base ten (we use ten different digits, 0 to 9, corresponding t our fingers, as the word "digit" suggests"). If we had 7 digits (0 to 6), for instance, the trick would work for 7-1=6 and its factors (2 & 3). Counting in base 7 is like this: 1 2 3 4 5 6 10(our 7) 11(our 8 ) 12(our 9) ...20(our 14)...100(our 49). Is 99 base 7 divisible by 6? Its digits add up to 9+9=24 (base seven!, our 18 ) which in turn add up to 6: so answer is yes. In fact 99 base 7 is our 48.

Would anyone like to prove the trick for dividing by 11 works?
WhistleTurning
 Quote: Is 99 base 7 divisible by 6? Its digits add up to 9+9=24 (base seven!, our 18 ) which in turn add up to 6: so answer is yes. In fact 99 base 7 is our 48.

Sorry this cannot exist as a 9 is an impossible digit in base 7, the maximum digit being 6???????
infinisa
Hi WhistleTurning

WhistleTurning wrote:
 Quote: Is 99 base 7 divisible by 6? Its digits add up to 9+9=24 (base seven!, our 18 ) which in turn add up to 6: so answer is yes. In fact 99 base 7 is our 48.

Sorry this cannot exist as a 9 is an impossible digit in base 7, the maximum digit being 6???????

Of course you are dead right - careless of me.

A better example, base 7, would be: Is 165 base 7 divisible by 6? Its digits add up to 1+6+5=15 (base seven!, our 12 ) which in turn add up to 6: so answer is yes. In fact 165 base 7 is our 1x49+6x7+5=96.

Thanks for spotting the error!
slashnburn99
ah nice posts,

if only the world was full of numbers instead of words it would be a better (16)(12)(1)(3)(5)
guissmo
 slashnburn99 wrote: ah nice posts, if only the world was full of numbers instead of words it would be a better (16)(12)(1)(3)(5)

i don't think it would. imagine trying to get used to words expressed in numbers (such as the one you said).
Fake
any tick with number nine is LAME

any multiple of 9 will always be a nice when you add the sum of the resultant numbers
add 1 to it, it wil be 10, and add TWO to it and add the result, it will always be 2!
chatrack
 friuser wrote: it's only a trick if you didn't go to school.

Yes, this is good trick by which we can make other amazed.
ciureanuc
I don't know English math terms so I will write only the formula:
the "trick" is useful when you ^2 numbers like 25 (35, 45, 75, etc.)

X<10
X5^2 = X(X+1) and add 25

example: 75^2 = 7x8 and write 25 after = 5625.
65x65 = 4225.
tukun2009manit
 chatrack wrote: Hai I have some cool trick with 9 ! step1. Immagine a number between 1 and 9 (single digit) step2. multilpy that number with 9 step3. If the result contain ONLY 1 digit continue to step 4. If result have 2 digits in it ADD the two digit step4. now add 1 to it step5. YOUR ANSWER IS 10 , hOW IS IT... COMMENT PLEASE

let me try to explain how this trick works, its simple to notice that in the multiple of 9 for eg:-
09
18
27
36
45
54
63
72
81
90
if you add the digits you will always get 9 for eg:-
0+9=9
1+8=9
2+7=9
.
.
.
.
.
.
.
.
8+1=9
9+0=9
this special factor of 9 is applied to make this trick

thus multiplying 9 with any no. b/w 0 to 9 gives u multiple of nine when the trick ask you to add the two digit which will be always 9 the extra addition of 1 is just to keep the trick transparent

you might be thinking is there any other no. other than 9 so that we can try this type of trick?
the answer is yes, try to notice the multiples of 11 you can see the here inspite of addition substraction leaves you with a constant

if you still have not understood pls write in the form i will try to clear it
rajpk
math ..........a science

nice topic

interesting replies

try to start topics like this......

thanx
sum12nv
not a very good trick.
try this one
type the bold numbers into a calculator(Dr x means multiply)
a woman was 69 years old but she was 222 fat. she wanted to look 51 years old. so she went to DR.X. He gave her 8 pills. She is now............(turn the calculator upside down!!!)
chatrack
Hi,

Here is another trick you can play with calculator.

step1. press 12345679 in the keypad [there is no 8]

step2. press x [multiply option]

step3. Ask your friend, to tell a single digit. If he said , say 7. Then 7x9=63

step4. press 63 and press "=" to get the result

Got the result 7777777 wow

ps: You can try this with any digit, except 0

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Find exact location of chat mate in any LIVE CHATROOM YM, gtalk, MSN,....

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guissmo
 chatrack wrote: step3. If the result contain ONLY 1 digit continue to step 4. If result have 2 digits in it ADD the two digit

You can replace this with "get the sum of all the digits" so that your "trick" becomes sneakier.
chatrack
guissmo wrote:
 chatrack wrote: step3. If the result contain ONLY 1 digit continue to step 4. If result have 2 digits in it ADD the two digit

You can replace this with "get the sum of all the digits" so that your "trick" becomes sneakier.

It may can bring a doubt how to add, so I preferred so, thanks for comment
Afaceinthematrix
 _AVG_ wrote: Yes ... but it is a divisibility test of 9 ... all multiples of nine have digits that add up to nine. Hence, if you add the digits and then add 1 to them, it will always come up to be 10. Similarly, all multiples of 3 have digits that add up to 3.

Actually, no. What you said is wrong. However, I am sure that you knew what you should say. All multiples of nine have digits that add up to a multiple of nine. They don't actually have to add up to nine. Similarly, all multiples of three have digits that add up to a multiple of three; they do not necessarily have to add up to three.

This is the reason why the OP was careful to only allow you to choose digits between one and nine. He/she could expand this to include all digits, but the OP would have to be careful and add several more steps to complete this.

That trick with the 7777777 is a little more interesting than the first trick and I encourage you to try and prove it. It shouldn't be too difficult to prove. Just play around with arbitrary numbers and see what you can get...
meep
I know that one it pained me for a while until i tried to do it the other way around.
Instead of typing in 12345679 * my number * 9, I did it the other way around and saw that 12345679 * 9 = 111111111.
So whatever number between 1 and 9 you choose you'll always get yournum * 111111111 which is why you'll always get your number 9 times.
Flakky
That was cool, simple trick. Didn't know that
tazone
another 9 trick

divide by 9

3/9=0.333...
30/99=0.303030...
25/999=025/999=0.025025025...

9/9=0.999...=1
james23
How it works?

Add step 2 & 4, you will get the right result.

In this case, you multiply 9 @ step 2 and add 1 @ step 4 and adding step 2 & 4, you get 10 as the result.
chatrack
 james23 wrote: How it works?

Any number X 9 will give out a number, whos digits on adding together again will get 9

Why this ? A math person can tell this
therimalaya

1. Imagine a three digit Number
2. Reverse the Numbers Digit
3. Subtract the smaller from bigger
4. Again Reverse the number of the result obtained in step 3
5. Add the reversed result and non reversed result in step 4
6. The result will always be 1089

nepalstar
 chatrack wrote: Hai I have some cool trick with 9 ! step1. Immagine a number between 1 and 9 (single digit) step2. multilpy that number with 9 step3. If the result contain ONLY 1 digit continue to step 4. If result have 2 digits in it ADD the two digit step4. now add 1 to it step5. YOUR ANSWER IS 10 , hOW IS IT... COMMENT PLEASE

Wow...! Great...! I never realized that...! Great...!
_AVG_
 therimalaya wrote: 1. Imagine a three digit Number 2. Reverse the Numbers Digit 3. Subtract the smaller from bigger 4. Again Reverse the number of the result obtained in step 3 5. Add the reversed result and non reversed result in step 4 6. The result will always be 1089 How about this 1089 trick...

This is a good one. However, it fails if you choose a palindrome 3-digit number. Try it for 131, for example! You get 0. So, I'd change Step 1 to "Imagine a non-palindrome 3-digit number".

Nevertheless, it can be deciphered. Here goes:
- by definition, you have a number 100a + 10b + c (a is not 0, a is not equal to c)
- Its reverse: 100c + 10b + a
- Difference: abs(100(c-a)-(c-a)) = 99(abs(a-c)) = 99k (where 'k' is positive, and of course, 1 < k < 9)
- 99k = 100k - k = 100k -100 + 100 - k = 100(k-1) + 10(9) + (10-k)
- Reversing the above, we have: 100(10-k) + 10(9) + (k-1)
- Therefore, the sum will be: 100(9) + 10(18 ) + 9 = 1089 !!

metalfreek
Keep on posting guys I might be a magician among my friends. I am learning all these tricks.