Sarah, a fifth-grade student, discovered that to test a number for divisibility

Question

Sarah, a fifth-grade student, discovered that to test a number for divisibility by 8, multiply the digits to the left of the units digit by 2 and add the result to the units digit. If the resulting number is divisible by 8 then the original number is divisible by 8. For example, for 336, the test is 2 times 33 + 6 = 72, so 336 is divisible by 8 because 72 is divisible by 8. Use diagrams of base-ten number pieces to provide an explanation that would make sense to a middle school student as to why this method works.

Explanation / Answer

1) Let us start with the simplest divisibility test--divisibility by 2, as the same idea carries over for 4 and 8.

If an integer has single digit x , it is clear that x is divisible by 2 iff x is 0,2,4,6,8!.

If an integer has more than one digit , write it as yx, where x is the unit digit and y is an integer

that has at least one digint.

the value of this integer is 10y +x (For example 59= 50 +9, 253 = 25x10 +3 and so on).

Now 10y +x is divisible by 2 if and only if x iis divisible by 2. (as 2 divides 10, irrespective of y).

From this we conclude that the divisibility of an integer (however long it may be) depends only on the unit digint x.

The number is divisibile by 2 iff the unit digit is divisible by 2.

2) Now consider 4. It is no longer enough to consider only the unit digit.-for example 12 adn 16 are divisible by 4 but their unit digits are not divisible by 4!.

But given any integer (assume it has more than 2 digits) can be written as

100z +(10y+x) . Now 4 always divides 100z irrespective of z (note however 4 need not always divide 10y!).

So 4 divides a number of the form 100z +(10y+x) if and only if 10y+x is divisible by 4.

Thus the divisiblility test for 4: Find out if the number formed by the last two digits is divisible by 4.

3) It is clear we need to include the number formed by the last 3 digits for divisibility by 8--consider the numbers 108, 116. The unit digit of the first and the number formed by the last two digits of the second are divisible by 8, but the numbers themselves are not divisible by 8.

Now assume the given integer has at least three digits.

Write the number as wzyx.

Consider the number 1000w + 2(100z+10y) +x. = 1000w+ 200z+ 20y+x. The first two terms are always divisible by 8 . So the number is divisible iff 20y+x is divisible by 8, as claimed in the question.

This proves the required result..

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