MSC #15: By Addition and By Subtraction

“Given that 3 x 37 = 111, what is the remainder when 222 222 222 237 is divided by 37?

1. 0; B) 1;  C) 4; D) 15;  E) 22”

This question, which was given in the Junior Level of the 2^nd International Vedic Mathematics Olympiad (IVMO 2022), can be answered immediately by using our 15^th of 30 MATH-Inic Specials for Christmas.

Grade school children are required to know by heart the divisibility rules for numbers such as 2, 3, 4, 5, 7, 8, 9, 10 and 11.  

A non-conventional, but logical method, to determine the divisibility of a number is by addition or subtraction of any multiple of the divisor.

In determining divisibility by any number, two facts must be remembered:

1. Addition of zeroes to and subtraction of zeroes from the end of a number does not affect its divisibility by a divisor.
2. Addition or subtraction of the divisor or a multiple of a divisor does not affect the divisibility of that number.

Is 1393 divisible by 7? By just adding 7 to 1393 we get 1,400 which obviously is divisible by 7. So, 1393 is divisible by 7.

Is 2346 divisible by 4? Since 100 is divisible by 4, we can deduct 2300 (23 x100) from 2346 to get 46. Then we can subtract 40 from 46 and get 6 which is obviously not divisible by 4. Thus, 2,346 is not divisible by 4.

By using number splitting (see MSC #3), in the IVMO question, we can easily see that  the number 222 | 222 |222 | 222 is divisible by 111 and consequently, by 37. And “By Subtraction” of this number from 222 222 222 237 , we will get 15, the remainder.  

Is 47, 492 divisible by 31? We repeatedly subtracted multiples of 31 from 47, 492 until we are left with 31, 000 which is obviously, divisible by 31.

The general idea is to produce zeroes at the end of the dividend, so let us list down the steps taken:

1. The dividend ends in 2 while our divisor ends in 1. To create a zero at the end, we multiply the divisor, 31, by the last digit of the dividend and deduct this product from the dividend: 47,492 – (2 x 31) = 47, 430.
2. To create another zero at the end, we multiply 31 by 30, and again deduct it from the remaining number: 47,430 – (30 x 31) = 47, 430 – 930 = 46, 500.
3. The next step is to multiply 31 by 500 to get 15, 500 which when deducted from 46, 500 will give 31, 000 which is obviously divisible by 31.

We can simplify the procedure above by removing the ending zeroes at the end in each step.

1. Same as the procedure before, getting 47, 430.
2. Remove the ending zero and multiply 31 by 3 and then deduct the product from the remaining number. 4743 – (3 x 31) = 4743 -93 = 4650
3. Repeat step 3: 465 – (5 x 31) =  465 – 165 = 310.

A further modification of this method is to just use 3, instead of 31.

1. Remove the ending digit in 47, 492, multiply it by 3 and deduct the product from the remaining number: 4749 – (2 x 3) = 4743
2. Repeating the procedure above until we arrive at a more conclusive number, we have 474 – (3 x 3) = 465.
3. Then 46 – ( 5 x 3) = 31      

Could you see the similarity between the method we just used and the divisibility test for 7?