How Predictable Are You? Three Digit Magic Trick

How predictable are you? Pick a 3 digit number with 3 different digits. Now reverse the digits to get a new number. You now have two numbers. Subtract the smaller number from the larger number.
Now add up the digits of the result. Can I guess your number?

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[alloosworld]

Uncle, my grandpa was maths teacher in a school so he teached me this awesome trick but now he left last year but I forgot this trick so then I saw your video so thankyou for explaining this much easily . Anyway thanks . I was crying cause I remembered my grandpa..Tnks a lot.🙂🙂😄😪😔🤧🙁😁😁😁

[yaiirable]

We don't need to use a brute force solution here.
Given the fact we know that a > c and b = b,
The final step will always look like: (a- c - 1) + 9 + (10 + c - a) = 18
We know that the final term will need to borrow a 1 that will be carried over all the way from the hundereds columns. a's and c's cancel out.

[Tehom]

0:34 Let's follow this trick using algebra:

Pick a 3-digit number:

I pick x*100 + y*10 + z, and I'll write it as "xyz"

Reverse the digits:

zyx

Subtract the smaller number from the larger number:

xyz - zyx = (x-z) 0 (z-x)

And I need to fix it up as a decimal, using the fact that xyz is larger so x > z:

(x-z-1) 9 (10+z-x)

Add up the digits of the result:

x - z - 1 + 9 + 10 + z - x = 18

[ProProboscis]

At this point you may need to create a kids' version of your channel.

[BlinkLed]

I don't know why these simpler videos always get so many dislikes. I appreciate the mix of content on the channel.

[badrunna-im]

Letting the number be ABC, from ABC-CBA we see that B cancels out.
Starting with a base case where A=2 and C=1, the difference will equal 99. Adding n to A will add 100n to the difference but also subtract n from it. The result can only be shifted by (100-1)n, or 99n. 99 from the base case +- 99n is also a multiple of 99.
Since 99=100-1, a multiple of 99 can be expressed as 99+(100-1)n. This shows that, between 3-digit multiples of 99, any digit's addition in the hundreds place is canceled out by the subtraction in the ones place. Without carrying over, i.e. when the last digit's 0 (990), the tens digit stays constant. Thus the sum of the digits is constant and 'guessable' within the constraints given in the question.

[fredrikenetorp]

Didn't work.

I choose 0x4BA
0x4BA --> 0xAB4 --> 0x5FA --> 0x1E = 30
which is not equal to 18 nor 0x18.

[ihordrahushchak]

numbers=[]
for x in range(100, 1000):
if

if reversed_num>x:
subtracted=reversed_num-x
else:
subtracted=x-reversed_num
added=0
for i in list(str(subtracted)):
added+=int(i)
numbers.append(added)
print(set(numbers))

Output:
{18}

[wolfgangsullifire]

I like the explanation and how you can work to get the answer

[nugepxs]

776-677 = 99
9+9 = 18

422-224 = 198
1+9+8 = 18

is break the rule but still get 18

[michaelempeigne]

Make more of these math videos please. Mindyourdecisions

[Packerfan]

Paused the video. Let 100a + 10b + c be your orginal number where 0 <= a, b, c <= 9 and each are distinct. Then the reverse number is 100c + 10b + a. The difference between them is |(100a + 10b + c) - (100c + 10b + a)| = |99a - 99c| = 99*|a - c|.
Then let 100x + 10y + z = 99*|a - c|. We need to show that x + y + z = 18. Now since 9 | (100x + 10y + z) and 11|(100x + 10y + z), we see that 9 | (x + y + z) and 11 | (x - y + z). Since 3 <= x + y + z <= 24 we see that x + y + z = 9 or 18. Since -8 <= x - y + z <= 17, we see that x - y + z = 0 or 11. Now suppose x - y + z = 11. Then x + z = y + 11. Then x + y + z = 2y + 11. Either 2y + 11 = 9 or 2y + 11 = 18. Thus, Either y = -1 or y = 3.5 both of which are impossible. Thus, x - y + z = 0. Thus, y = x + z. Thus, x + y + z = 2y. Either 2y = 9 or 2y = 18. Thus, either y = 4.5 or y = 9. However, y = 4.5 is impossible. Thus, y = 9 and x + z = 9. Thus, x + y + z = 18.

[AndreasWoykepianistandcomposer]

Just to add: The result of each one digit number multiplied by 99 will be a three digit number with a 9 in the middle. The other two digits are supposed to add up to 9 because the checksum of each number divisible by 9 is also divisible by 9. So the result will always be 18. Like it!

[okaro]

The first number of the result is a-c-1, the second 9 and the last 10+c-a.  Add these and the variables cancel each other and you get -1+9+10 = 18.

[fkfk]

I thought this was a different problem. Take a three digit number as stated, and subtract. Then take the result and reverse it again and add the two numbers. The number always is 1089, and I am not sure why.

[TheOriginalFayari]

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[antforeargar]

Other solution:
abc
- cba


a0c
- c0a


There will be a 9 in the middle because a>c

The Numbers will now be: a-c-1 (-1 because you need to borrow so that you can borrow from the middle number), 10-1, 10-(a-c) (the 10 is from the middle number, and c-a = -(a-c) (not actually necessary, but That's how I solved it))

Now we add all of the numbers.
a-c-1 + 9 + 10-(a-c) = a-c-1 +9 + 10+c-a = a-a+c-c+10+9-1= 18
Answer will always be = 18

(This Was all written on a mobile device)

[x_abyss]

I used similar approach up until checking if every possibility sums up to 18. We know that the subtraction of the larger number from the smaller one is a multiple of 99, which means all the digits of the result add up to a multiples of 9, and the sum of the 1st and 3rd digits minus the 2nd is a product of 11. The only way the latter criteria can be satisfied is if 1st digit + 3rd digit - 2nd digit equals 0, meaning that the sum of the difference is even. Therefore the only sum of any three digits number divisible by both 2 and 9 are either 0 or 18.

[nugepxs]

If I break the some rule
It can also work some same digit with first digit and second digit
Or same in second digit and last digit

It only didnt work is first digit and last digit is same number

644-466 = 198
1+9+8 = 18

But if 464 - 464 = 0

You can change "with 3 different digit" to "with first digit and last digit is different number"

[saraflint]

I knew the trick as soon as you said "reverse the digits and subtract." This also works with 4-digit numbers. With 5 and 6-digit numbers, the end sum is 27. With double-digit numbers, the end sum is 9. (With single-digit numbers, the end sum is also 9. Example: 70 - 07 = 63. 6 + 3 = 9.) Good trick for beginners, but I'm onto you now!