The Missing Digit Trick !

[Mental Math]

Ask a friend to secretly write down ANY number (at least four digits long). e.g. 78341

Ask the friend to add up the digits... e.g. 7+8+3+4+1 = 23 ... and then subtract the answer from the first number. e.g. 78341 - 23 = 78318

Your friend then crosses out ONE digit from the answer. (It can be any digit except a zero) e.g. 7x318

Your friend then reads out what digits are left .e.g. 7-3-1-8 Even though you haven't seen any numbers, you can say what the missing digit is! EIGHT

THE SECRET This great trick relies on the power of 9. How? Link removed by Moderator

Edited February 8, 2012 by Phi for All
Advertising link removed

[Schrödinger's hat]

Ask a friend to secretly write down ANY number (at least four digits long). e.g. 78341

Ask the friend to add up the digits... e.g. 7+8+3+4+1 = 23 ... and then subtract the answer from the first number. e.g. 78341 - 23 = 78318

Your friend then crosses out ONE digit from the answer. (It can be any digit except a zero) e.g. 7x318

Your friend then reads out what digits are left .e.g. 7-3-1-8 Even though you haven't seen any numbers, you can say what the missing digit is! EIGHT

THE SECRET This great trick relies on the power of 9. How? Link removed by Moderator

 

Seeing as this was advertising and the link was removed. For anyone wondering how it's done:

 

Adding the digits of a number results in a number with the same value modulo 9

 

So 7 + 8 + 3 + 4 + 1 is 5 (mod 9)

Subtract the sum of digits, 23, which is also 5 (mod 9)

This guarantees you have a number that is 0 (mod 9)

 

So from the sum of digits having the same value mod 9 property, we know that if we add the digits we'll get a number that is 0 mod 9.

Erasing one digit reduces that value by the value of that digit.

So sum the digits that are read out, 9 - sum is the value of the removed digit.

[DrRocket]

Seeing as this was advertising and the link was removed. For anyone wondering how it's done:

 

Adding the digits of a number results in a number with the same value modulo 9

 

So 7 + 8 + 3 + 4 + 1 is 5 (mod 9)

Subtract the sum of digits, 23, which is also 5 (mod 9)

This guarantees you have a number that is 0 (mod 9)

 

So from the sum of digits having the same value mod 9 property, we know that if we add the digits we'll get a number that is 0 mod 9.

Erasing one digit reduces that value by the value of that digit.

So sum the digits that are read out, 9 - sum is the value of the removed digit.

 

Proof: 10=1 (mod 9)

[TonyMcC]

Just out of interest this is something I posted a couple of years ago which uses the same argument to show that the square root of 2 must be infinitely long.

 

 

Many years ago I was shown a simple check for multiplication.

The check can indicate a definite mistake; but cannot indicate definite accuracy.

The check is performed by adding the individual digits of each part of the calculation. If this results in a number with more than one digit then add the digits of the new number and repeat as necessary until you have a single digit representing each part of the calculation.

Example:-

 

39 leads to 3+9=12, and 1+2=3

58 leads to 5+8=13, and 1+3=4

 

39 x 58=2262 leads to 2+2+6+2=12, and 1+2=3

 

If the single digit representing 39 (i.e.3) is multiplied by the single digit representing 58 (i.e.4) we get 12 which leads to 1+2=3. The fact that this agrees with the single digit representing the answer indicates that the calculation might be correct, but more importantly if the single digit representing the answer had been any other number the calculation would have been performed incorrectly.

 

Another example:-

 

58764 (5+8+7+6+4=30 ; 3+0=3)

3271 (3+2+7+1=13 ; 1+3=4)

------------

192217044 (1+9+2+2+1+7+0+4+4=30 ; 3+0=3)

 

Performing the check, 3 x 4=12 and 1+2=3

since this agrees with the answers single digit no error has been detected.

 

Although this description might seem complicated it is really very simple and can be done "in your head" with very little practice.

 

Examining the square root 0f 2

_________________________

 

Think of the calculation A x A= 2

So A must be the square root of 2

 

Someone suggests that the square root of 2 is 1.4

If they are correct then 14 x 14 =200

 

So using the test

14 (1+4=5)

14 ( 1+4=5) and 5x5 = 25, and 2+5=7

----

200 (2+0+0=2)

Since "7" and "2" do not have the same value this calculation must be incorrect.

 

Someone then suggests 1.41

 

So using the test

141 (1+4+1=6)

141 (1+4+1=6) and 6x6=36, and 3+6=9

---------

20000 (2+0+0+0+0 =2)

Since"9" is not "2" this calculation is incorrect.

 

My calculator gives the square root of 2 as 1.414213562

therefore 1414213562^2=2 x 10^18.

If you do the test the two relevant digits are "4" and "2" indicating an error.

 

The fact is you can add digits beyond 1.414213562 all the way to infinity without getting a "correct" indication!

 

This argument holds good for many (probably most) numbers.

 

Perhaps any whole number falls in to one of two categories. Perhaps it either has an easily found whole number square root or it is an infinitely long decimal number. Perhaps this is true of any number base you might use (octal, hexadecimal etc.)??

This post has been edited by TonyMcC: 22 February 2010 - 10:30 PM

[DrRocket]

Just out of interest this is something I posted a couple of years ago which uses the same argument to show that the square root of 2 must be infinitely long.

 

Proof:

 

Any finite decimal represents a rational number.

 

[math]\sqrt 2 [/math] is not rational.

 

QED

 

 

 

Proof that [math]\sqrt 2 [/math] is not rational.:

 

If it were then [math]\sqrt 2 = \frac {p}{q} [/math] where[math] p [/math] and [math] q [/math] are relatively prime. So.

 

[math] 2 = \frac {p^2}{q^2}[/math]

[math] 2q^2 = p^2[/math]

 

Therefore [math'p[/math] is even and [math]2q^2 = 4s^s[/math] for some [math]s[/math]. But then [math]q[/math] is also even, contradicting the fact that [math]p[/math] and [math]q[/math] are relatively prime.

 

QED

[TonyMcC]

Proof:

 

Any finite decimal represents a rational number.

 

[math]\sqrt 2 [/math] is not rational.

 

QED

 

 

 

Proof that [math]\sqrt 2 [/math] is not rational.:

 

If it were then [math]\sqrt 2 = \frac {p}{q} [/math] where[math] p [/math] and [math] q [/math] are relatively prime. So.

 

[math] 2 = \frac {p^2}{q^2}[/math]

[math] 2q^2 = p^2[/math]

 

Therefore [math'p[/math] is even and [math]2q^2 = 4s^s[/math] for some [math]s[/math]. But then [math]q[/math] is also even, contradicting the fact that [math]p[/math] and [math]q[/math] are relatively prime.

 

QED

 

This is copy of a post I made later in my earlier thread:-

 

TonyMcC

 

Quark

Thank you "the tree". If you look at my earlier entries for this subject you will see that you seem to have answered my earlier supposition. However I still am not really happy with your explanation concerning irrationality. As I understand it the square root of 2 is accepted as irrational not because its decimal expansion extends to infinity but for a completely different reason. It is easily proved that when expressed as a fraction and in its most simplified form both parts of the fraction must be even numbers. That really is irrational! Is it a proven fact that all square roots of whole numbers are either whole numbers or irrational? How is the irrationality proved?

 

DrRocket - Glad you agree about why the square root of two is considered irrational.

Edited February 10, 2012 by TonyMcC

[DrRocket]

Is it a proven fact that all square roots of whole numbers are either whole numbers or irrational? How is the irrationality proved?

 

Let [math]n[/math] be a positive integer. Suppose that [math] \sqrt n = \frac {p}{q}[/math] where [math]p[/math] and [math]q \ne 1[/math] are relatively prime positive integers.

 

Then [math]n^2 = \frac {p}{q}[/math] and hence [math]p=n^2q[/math] contradicting the assumption that [math] p[/math] and [math]q [/math] are relatively prime.

 

QED

 

Note that this is really just a slight generalization, with reduced verbage, of the argument that I gave earlier for [math]\sqrt 2 [/math] being irrational.

 

 

 

DrRocket - Glad you agree about why the square root of two is considered irrational.

 

I did not agree on anything. The square root of two is considered irrational because it IS irrational.

[TonyMcC]

 

I did not agree on anything. The square root of two is considered irrational because it IS irrational.

 

Surely you can't say that unless you can prove it ? Are you saying it has never been proved?

Edited February 11, 2012 by TonyMcC

[DrRocket]

Surely you can't say that unless you can prove it ? Are you saying it has never been proved?

 

Can you read ?

 

I did prove it. Twice.

Edited February 11, 2012 by DrRocket

[TonyMcC]

Can you read ?

 

I did prove it. Twice.

 

I thought that what you proved was justifying what I said in #6 above. "It is easily proved that when expressed as a fraction and in its most simplified form both parts of the fraction must be even numbers." (Copied from "Square root of 2: #8: 30th Apr 2010)

 

If I misunderstood then I apologise and would appreciate you pointing out where we differ. I would appreciate you doing so without the use of sarcasm.

Edited February 11, 2012 by TonyMcC

[DrRocket]

If I misunderstood then I apologise and would appreciate you pointing out where we differ. I would appreciate you doing so without the use of sarcasm.

 

Read.

 

BTW I may have reproduced one argument with which you are familiar. That is not surprising as the argument is over two thousand years old -- generally attributed to Pythagorus. But I most certainly did not and have no need to copy such an elementary argument.

Edited February 12, 2012 by DrRocket

[TonyMcC]

Read.

 

BTW I may have reproduced one argument with which you are familiar. That is not surprising as the argument is over two thousand years old -- generally attributed to Pythagorus. But I most certainly did not and have no need to copy such an elementary argument.

 

Thank you.

 

BTW I would hardly claim the proof is mine since, as you will know, it is one of the first illustrations of the concept, irrationality, anyone starting to understand it gets shown.

 

Everything copied and pasted in this thread is a repeat of my own work and done for complete accuracy should anyone want to verify that as a fact.

 

I would like to end by saying clearly that I appreciate your time given and respect the depth of knowledge you have in your subject.

Edited February 12, 2012 by TonyMcC