Given a number, the task is to quickly check if the number is divisible by 17 or not.
Input : x = 34 Output : Yes Input : x = 47 Output : No
A solution to the problem is to extract the last digit and subtract 5 times of last digit from remaining number and repeat this process until a two digit number is obtained. If the obtained two digit number is divisible by 17, then the given number is divisible by 17.
- Extract the last digit of the number/truncated number every time
- Substract 5*(last digit of the previous number) from the truncated number
- Repeat the above three steps as long as necessary.
3978-->397-5*8=357-->35-5*7=0. So 3978 is divisible by 17.
Mathematical Proof :
Let be any number such that =100a+10b+c .
Now assume that is divisible by 17. Then
0 (mod 17)
100a+10b+c 0 (mod 17)
10(10a+b)+c 0 (mod 17)
10+c 0 (mod 17)
Now that we have separated the last digit from the number, we have to find a way to use it.
Make the coefficient of 1.
In other words, we have to find an integer such that n such that 10n1 mod 17.
It can be observed that the smallest n which satisfies this property is -5 as -501 mod 17.
Now we can multiply the original equation 10+c 0 (mod 17)
by -5 and simplify it:
-50-5c 0 (mod 17)
-5c 0 (mod 17)
We have found out that if 0 (mod 17) then,
-5c 0 (mod 17).
In other words, to check if a 3-digit number is divisible by 17,
we can just remove the last digit, multiply it by 5,
and then subtract it from the rest of the two digits.
Note that the above program may not make a lot of sense as could simply do n % 23 to check for divisibility. The idea of this program is to validate the concept. Also, this might be an efficient approach if input number is large and given as string.
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