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;
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:
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].
we deduce that;
and if [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.
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]
then it is also correct for [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.
If n is an integer, then [pic] is an integer implies that [pic] is divisible by 5. Therefore,
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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