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I wish to explain to a grade 5 kid, the answer to the question - 'When can we know that the observing the last digit of a number will not be sufficient to determine its divisibility by a given N?'

For example - It doesn't work for 3, because we get all sorts of digits, i.e. from 0 to 9, at the units place. The same reason works for 7 and 9.

But what would be a convincing argument for 4 and 8? Even though the last digits form a pattern - 0, 2, 4, 6 and 8 - but we're still not sure about its divisibility.

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  • $\begingroup$ $4\cdot 25 = 100$, so the last $2$ digits are all that matters. $\endgroup$ – Eric Lee May 26 '17 at 5:26
  • $\begingroup$ Looking at the last two digits works for four, and similarly three digits for eight. $\endgroup$ – Parcly Taxel May 26 '17 at 5:26
  • $\begingroup$ @EricLee I understand that. But the question is something else. To explain divisibility by 2, most books proceed like "Observe the multiples of 2 - 2, 4, 6, 8, 10, 12, 14 ... Observe their last digits. Hence all numbers ending in 2, 4, 6, 8, 0 are divisible by 2". But this doesn't work for 4 suddenly. For 3, we could argue that it doesn't work because we get all possible digits at the units place. But what about 4? $\endgroup$ – yomayne May 26 '17 at 7:02
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Any positive integer $N$ can be written as $$100a+10b+c$$

$$=4(25a+2b)+2b+c$$

So, we need $2b+c$ to be divisible by $4\iff N$ is to be divisible by $4$

Another observation: Clearly $c$ must be even $=2d$(say)

So, we need $b+d$ to be even i.e., $b,d$ must of same parity i.e., either both even or both odd

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I think we have use example to explain this to a Grade 5 kid.

$$2316=23\times 100+16=23\times (25\times 4)+16=(23\times 25)\times 4+16$$

$16$ is divisible by $4$ because $16=4\times 4$. This also means that

$$2316=(23\times 25)\times 4+4\times 4=(23\times 25+4)\times 4$$

and it is divisible by $4$.

On the other hand,

$$2317=(23\times 25)\times 4+17$$

$17$ is not divisible by $4$ since $17=4\times 4+1$ (we have a non-zero remainder). This also means that

$$2317=(23\times 25)\times 4+4\times 4+1=(23\times 25+4)\times 4+1$$

and it is not divisible by $4$, as we have a non-zero remainder.

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To understand how the divisibility test for $4$ works it's helpful to think about how we write numbers, specifically in terms of a base $10$ expansion. For example, when we write something like $1234$ what we really mean is $$1000 + 200 + 30 + 4.$$ Everything past the $10$'s place is automatically divisible by $4$, because $100$, $1000$, etc. are all divisible by $4.$ The $10$'s place and the $1$'s place are where we need to check, that is we need to check if the last two digits are divisible by $4$ because that is the part of the number that is not automatically divisible by $4$ when thinking about it in terms of a base $10$ expansion. In this example $1234$ is not divisible by $4$ because while both $1000$ and $200$ are divisible by $4$, $34$ is not. On the other hand $1236$ would be divisible by $4$ because $36$ is.

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