Use the reasoning for the divisibility test for 3 and the reasoning about the re

Question

Use the reasoning for the divisibility test for 3 and the reasoning about the remainder when dividing by 9 to devise a divisibility test for 9. Sometimes children are taught a method for checking arithmetic calculations called costing our nines. Perhaps you know it. This method is easy enough for an upper elementary student to use, but understanding why it works involves some of the notions we've been discussing. Here's how it works if you wish to check for errors in addition: 326 479 +84/889 Step 1: For each number, cross out ("cast out") digits that are 9 or whose sum is 9. Step 2: Add the remaining digits until you have a number 0-8. This will be your "reduced number." Step 3: Do the operation indicated on the reduced numbers. Step 4: Check to see if the sum. difference, or product, as appropriate, of the reduced numbers matches the reduced number of the sum. difference, or product. If it doesn't, check further for an error. For the above addition problem, here is an illustration of the method:/2/leftarrow 2 (Also, when 326 is divided by 9. the remainder is 2.) 47/leftarrow 4 + 7= 11; 1 + 1 = 2 (Also, when 479 is divided by 9, the remainder is 2. + 84 leftarrow 8 + 4 = 12; 1 + 2 = 3 (Also, when 84 is divided by 9, the remainder is 3.) 88/leftarrow 8 + 8= 16; 1 + 6 = 7 (Also when 889 is divided by 9, the remainder is 7.) If the sum is correct, then the sum of 2, 2, and 3 should equal the reduced number of the original sum. It does! 2 + 2 + 3 = 7. Why does this work? It is based on the fact that the sum of the digits of a number is the remainder when dividing by 9. Consider: 326 = 9 times 36 + 2 (Note that 3 + 2 + 6 = 11 and 1 + 1 = 2.) 479 = 9 times 53 + 2 (Note that 4 + 7 + 9 = 20 and 2 + 0 = 2.) + 84 = 9 times 9 + 3 (Note that 8 + 4 = 12 and 1 +2 = 3.) 889 The number 889 should equal 9(36 + 53 + 9) + (2 + 2 + 3) = 9(98) + 7. and it does. A similar technique works for checking products.

Explanation / Answer

Idivisibility test for 9:

A number is divisible by 9 if and only if the sum of the digits of that number is divisible by 9.

This is because, the sum of the digits is same as the remainder obtained when dividing the number by 9.

If its a multiple of nine then it can be taken as zero.

The similar result works for 3 as well:

A number is divisible by 3 if and only if sum of the digits is divisible by 3

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