The divisibility rule for 9 can be extended for divisors composed entirely of 9s like 99, 999, 9999 etc.
The rule for 9 is that “If the sum of the digits of the number is divisible by 9, then the number is divisible by 9”. We can avoid division by repeating the addition if the sum of the digits has more than 1 digit. For example, for the number 4851, the sum of the digits is 18, so we add 1 and 8 and get 9. Thus, the number is exactly divisible by 9.
Now if we add 2 to 4851, we get 4853. This number obviously would have a remainder of 2 when divided by 9. Let us see what the digit sum of 4853 is. Adding the digits will give us a total of 20 and 2 + 0 = 2. We can see here that the digit sum is the remainder when a number is divided by 9.
“Casting out 9s” is a technique to quickly find the digit sum of a number. We can cast out or delete a 9, or digits totaling 9 or a multiple of 9. For 4851, we can cast out 4+5 and 8 +1, leaving zero.
Common divisibility tests will be discussed by IAVM Trustee Gowri Ramachandran on Dec 5 while Osculation, a systematic divisibility test for any number will be explained by MATH-Inic Training Director Veronica S. Prudente on Dec 11, 2021, in the Inspirational Maths from India (Year 4) webinars. (Please send a message to “Ike Prudente” on Messenger or to https://www.facebook.com/MATHInicPhils for registration details.)
In our example, to find out if 48, 919, 032 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: 032 + 919 + 48 = 999. The number is then divisible by 999.
How about 995,798, 897? If we solve 897 + 798 + 995 we will get2690. Then we add again 690 + 2, to get 692. We can say that 995,798,897 will have a remainder of 692 when divided by 999.
Try to figure out how if “casting out” 999 can be applied in this case.
Can we also develop divisibility rules for numbers like 299, 5001, etc?