Tell A Number Trick
by Paul Chika Emekwulu, Author  "Magic of Numbers"
"Let's play a game," Lori said to her
friend, Tessi. Both are students at Bahome High School, North  East Potomico.
"What kind of game is that?" Tessi asked in return.
"What you need is just a piece of paper and a pencil," Lori continued.
Lori and Tessi have started the game. Below are
the instructions:
 Choose any number with two or more digits, e.g.
467896
 Add the digits, i.e. ( 4 + 6 + 7 + 8 + 9 + 6
) = 40
 Subtract this sum from the original number, i.e.467896
 40 = 467856
 Cross out any digit you like and read out the
remaining digits in order: 4 6  7  8 5 (gradually).
Now, how do we know the number Tessi crossed out?
Do you want to know? Now listen and make sure you follow the investigation
carefully. Promise to go diligently with me through this investigation
and find out the number Tessi crossed out.
Now, the algebraic investigation
 Let's take a number, say 54. By adding the digits
we have: ( 5 + 4 ) = 9
 Subtract this sum from the original number, i.e. 54  9 = 45
Let's take more numbers, e.g. 74. Then 7 + 4 = 11
and 74  11 = 63.
Another number, say 87. 8 + 7 = 15 and 87 
15 = 72
At this stage, it seems that the number left after subtraction
is always divisible by 9. We don't know yet. It is only an assumption at
this point. What do you think? If the above assumption is true, it means
that such a number will beautifully obey the divisibility rule of 9.
Back to our example, 54.
 Adding the digits in 54, we have 5 + 4 = 9.
 Now, let's cross out any digit in 54. Which do you want crossed out?
It doesn't make any difference. Crossing out 4, we are now left with 5.
At this point of the investigation, let's go back to the divisibility by
9 and get ourselves acquainted once more with it. This test states that
a number is divisible by 9 if the sum of its digits is divisible by 9.
Do you still remember this rule?
 In our example, since the number 5 left after crossing out 4 is only
a one digit number, the number left after the procedural mental addition
is 5. But let me ask you a question.
If in the number left and after subtracting the sum of digits, I cross
out a digit, how else do you think we can easily find the number crossed
out other than by applying the divisibility rule of 9? Think about this.
 Now let us represent 4 in 54 with m. Since 54 is divisible
by 9, then, 5 + m (sum of digits) is also divisible by 9.
At this point, what do you think about the crossed out digit?
Now, listen. What number should be added to 5 such that the resulting number
is divisible by 9? Think of 4, 13, 22, 31, 40... as possibilities.
Again, remember that the number crossed out is a one digit number. From
the above series, what is the number satisfying our condition for the values
which m can take?
The answer is 4.
Therefore, the smallest value m can take is 4 so that 5 + m
will be divisible by 9.
How? From the above you will see that:
5 + 4 = 9
5 + 13 = 18
5 + 22 = 27
5 + 31 = 36
5 + 40 = 45
These sums are all multiples of 9. From the above, the highest multiple
of 9 with respect to 5 is 9. Therefore, m (number crossed
out) = 9  5 = 4.
Now, let's have a general look at the situation. Consider a number,
abcd, whose other name is 1000a + 100b + 10c + d.
This expression could be expressed further as:
( 999a + a ) + ( 99b + b ) + ( 9c + c ) + d
This last expression in turn, also is equal to:
( 999a + 99b + 9c ) + ( a + b + c + d ) = 9 ( 111a + 11b + c )
+ ( a + b + c + d )
The investigation continues.
a + b + c + d represents the sum of digits in the original
number, abcd. Therefore, subtracting this sum from the original
number, which is equal to:
9 ( 111a + 11B + c ) + ( a + b + c + d )
we have,
9 ( 111a + 11b + c ) + ( a + b + c + d )  a  b  c  d = 9 (
111a + 11b + c )
The expression, 9 ( 111a + 11b + c ) is a multiple of
9.
Do you agree? In other words, generally, the number left after subtracting
the sum of digits from the original number, abcd, is always
a multiple of 9. It is therefore, divisible by 9 since it has 9 as a factor.
Again, since 9 ( 111a + 11b + c ) = 1000a + 100b + 10c + d  (
a + b + c + d ), 1000a + 100b + 10c + d  ( a + b + c + d
) is, consequently, a multiple of 9.
Again, 1000a + 100b + 10c + d  ( a + b + c + d ) = 1000a + 100b
+ 10c + ( d  d )  ( a + b + c ) = 1000a + 100b + 10c  ( p + q + r )
where p, q, r are the numbers
left after the crossing out process.
When the original number is abcd, the numbers left after
crossing out will be p, q, and r.
Do you agree with this logic?
Continuing our investigation, let 9k be a multiple of
9 where k is determined by the quantity,
Therefore, the number crossed out is the expression, 9k  ( P
+ q + r ).
 Let's make an illustration with this example by choosing the number,
6345 as our original number.
 Adding the digits we have: 6 + 3 + 4 + 5 = 18
 Subtracting the sum of difits from the original number we have: 6345
 18 = 6327
 Let's cross out 6.
Then, we have 327 left.
Therefore, p = 3 , q = 2 , and r = 7
 Adding p, q, and r, we have:
3 + 2 + 7 = 12
 The next highest multiple of 9 more than 12 is 18. i.e. k = 2
 Therefore, number crossed out = 9k  ( p + q + r ).
 By substitution, m = 18  ( 3 + 2 + 7 ) = 18  12 = 6
Experience has shown that students enjoy this trick. At all of my presentations,
they always get excited and are always very anxious to understand the algebra
explaining the trick.
An adaptation from Magic of Numbers, Grades 912 by Paul Chika Emekwulu
Magic of Numbers, Grades 912 is available at the following: amazon.com, barnesandnoble.com, borders.com or through the publisher:
Novelty Books
P.O.Box 2482
Norman, OK 73070
Tel/Phone: (405) 447  9019
Magic of Numbers is downloadable at publishingonline.com
Be sure to read Paul Chika Emekwulu's chat August 19, 1998, this month featured live chat from the Teachers.Net Archive!
