Question 1

To test for divisibility by 11 in any three-digit number, we can look at the following general case:

Let [pic] be our three-digit number that is divisible by 11, then;

[pic]

Therefore if the alternating digit sum i.e. [pic] is divisible by 11 then the original three-digit number is also divisible by 11.

Similarly for a five-digit number:

[pic]

Therefore if the alternative digit sum i.e. [pic] is divisible by 11 then the original five-digit number is also divisible by 11.

If we consider an integer with more digits, let’s call it [pic] with digits

[pic] (units),   [pic] (tens),   [pic] (hundreds),   etc.

Then [pic]   where [pic] is the number of digits in [pic].

Now,
[pic]

we deduce that;

[pic]

and if [pic]

then [pic]
Question 1 contd.

Therefore if   [pic] is divisible by 11 then its alternating digit sum, starting with the units digit is also divisible by 11.

For example, with [pic]= 868,476,433

the alternating digit sum is

3 – 3 + 4 – 6 + 7 – 4 + 8 – 6 + 8 = 11

Hence 868,476,433 is divisible by 11.

Question 2

Proof 1:
If n is an integer, then [pic] is an integer implies that [pic] is divisible by 5.   Therefore, [pic] for some [pic].

Let [pic] be the statement ‘if n is an integer, then [pic] for some [pic]’

Firstly, we prove that [pic] is true for n = 1

P(1) = 0, therefore true.

Secondly, we assume [pic] to be true for n equal to some arbitrary [pic]

[pic]

then it is also correct for [pic]

[pic]

This is clearly divisibly by 5.   Since [pic],
then by the principle of mathematical induction [pic] is true and therefore if n is an integer, then [pic] is an integer.

Proof 2:

If n is an integer, then [pic] is an integer implies that [pic] is divisible by 5.   Therefore,
[pic]

Since we have factors [pic], we have three consecutive numbers and therefore any number n whose last digit is 0, 1, 4, 5, 6 or 9 will be divisible by 5.   When... [continues]

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