Question 1204005: A Spooky number is a number which is 13 times the sum of its digits.
Find all Spooky numbers (and prove that there aren’t any more).
Answer by math_tutor2020(3835)   (Show Source):
You can put this solution on YOUR website!
One digit numbers:
x = some positive integer in the set {0,1,2,3,...,8,9}
If x was "spooky", then x = 13x which solves to x = 0
Therefore, 0 is the only spooky 1 digit number.
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Two digit numbers:
Consider the 2 digit number 10a+b, where a,b are positive integers.
Eg: a = 2 and b = 3 gives 10a+b = 23.
If a spooky number could be 2 digits, then this equation would be true
10a+b = 13*(a+b)
Solve for 'a' to get a = -4b.
If b > 0, then a < 0.
This lets us conclude that there aren't any 2 digit spooky numbers that are positive.
Furthermore, a+b > 0 so 13*(a+b) > 0. This means spooky numbers cannot be negative.
Conclusion for this section: There aren't any spooky 2 digit numbers.
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Three digit numbers:
a,b,c are positive integers
100a+10b+c is some 3 digit number
100a+10b+c = 13*(a+b+c) must be the case if this number is considered "spooky".
Solve for 'b' to get b = 29a-4c
Plug that into 100a+10b+c to get...
100a+10b+c
100a+10(29a-4c)+c
100a+290a-40c+c
390a-39c
39*(10a-c)
This shows 100a+10b+c is a multiple of 39 if 100a+10b+c is spooky.
Let's look at 13*(a+b+c) more closely.
Specifically a+b+c
The smallest that sum can be is 1 because of the number 100
The largest is 9+9+9 = 27
1 <= a+b+c <= 27
or
1 <= a+b+c <= 9*3
Multiply both sides by 13 to get
1 <= a+b+c <= 27
13*1 <= 13*(a+b+c) <= 13*27
13 <= 13*(a+b+c) <= 351
It appears that 3 digit spooky numbers could be possible.
I'll let the student search for them.
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Four digit numbers:
We'll do a similar trick as the previous section.
1 <= a+b+c+d <= 9+9+9+9
1 <= a+b+c+d <= 9*4
1 <= a+b+c+d <= 36
13*1 <= 13*(a+b+c+d) <= 13*36
13 <= 13*(a+b+c+d) <= 468
We run into a problem. The largest 13*(a+b+c+d) can be is 468 which is clearly NOT a 4 digit number.
There's no way to have 13*(a+b+c+d) as a 4 digit number.
Therefore, we conclude that 4 digit spooky numbers aren't possible.
The logic for 5 digit numbers, 6 digits, and so on will follow the same idea. I'll let the student finish the proof.
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