 # MSC #16: Extending the Divisibility Rule for 9

“What is the remainder when 48519872 is divided by 9”

This question, given in Primary Level of the recently concluded 2nd International Vedic Mathematics Olympiad, can be easily answered by two methods as we explained two days ago in MSC# 14: Digit Sums – 1) by repeated digital sum or 2) by “casting out 9s”.

By adding all the digits of the number, we get 4 + 8 + 5 + 1 + 9 + 8 + 7 + 2 = 44 and if we deduct 36 which is a multiple of 9 from 44, we will get 8, the remainder. We can also get 8 if we add the digits of 44:  4 + 4 = 8.

We can get the same results, easier,  if we cast out 9s or digits adding up to 9s: we can disregard the leading 4 and the 3rd digit 5; the 2nd and 4th digits 8 and 1, the 5th digit 9; and the last two digits 7 and 2. This leaves only 8 which is the remainder we are looking for.

But what would be the remainder when the divisor is 99 or 999?

How does the divisibility rule for 9 works? Let’s take an example:

3, 231 = 3000 + 200 + 30 + 1

= 3(999 + 1) + 2(99 +1) + 3( 9 + 1) + 1

= [3(999) + 2(99) + 3(9)]  + (3 + 2 + 3 + 1)

= 9[3(111) + 2(11) + 3(1)] + (3 + 2 + 3 + 1)

Since the first part is clearly divisible by 9, the whole number’s divisibility by 9 is determined only by the second part which is the sum of the digits of the number.

Let us find out if we can extend this rule for 99 and 999.

254, 133 = 250 000 + 4100 + 33

=   25(9999 + 1) + 41 (99 +1) + 33

= [25 (9999) + 41(99)]  + (25 + 41 + 33)

= 99 [ 25(101) + 41(1)]  + (25 + 41 + 33)
= 99 [ 25(101) + 41(1)] + 99

Since the first addend is a multiple of 99 and the second addend is 99, we can say that 254, 133 is divisible by 99.

In our illustrated example, to find out if 2, 355, 642 is divisible by 999, we partition the number into groups of 3 digits each starting from the right. Then we will find the sum of these groups of 3 digits: 642 + 355 + 2 = 999. The number is then divisible by 999.

Now going back to the IVMO question, we can use the extended divisibility rules to find out the remainders when 48519872 is divided by 99 and 999 respectively.

If we partition the number into groups of two digits each starting from the right and get their total, we will have 72 + 98 + 51 + 48 = 269 and if we add again, we get  69 + 2 = 71. We can get this same remainder “by casting out 99s”. Note that 51 and 48 add up to 99 and thus, can be disregarded. Then we can deduct 1 from 72 and add it to 98 to make it 99. That leaves a remainder of 71.

If we group the digits by three from the right and add, we will get 872 + 519 + 48 = 1,439 and if we add again, we will get 439 + 1 = 440. We will get the same result if we subtract 999 from 1,439

Can we also develop divisibility rules for numbers like 299, 5001, etc?