Write an expression for the difference between the number...?

Question

The digits of a number are a,b,c.
i) Explain why the number may be written 100a + 10b + c
Because they're in the hundreds, tens and units columns.

ii) Write down an expression for the difference between the number and the sum of its digits.

?

iii) Prove that, if the sum of the digits is divisible by 3, then so is the number.

Would I have to give an example of a number e.g. 2+3+4 and 234 but that wouldn't really prove it?

Can someone help me answer ii) and iii)?
Thanks.

Answer 1

Well, you already have the expression for the number:

100a+10b+c

The sum of the digits is a+b+c, so the difference between the two is

100a+10b+c - (a+b+c), which being simplified is 99a + 9b

For #3, consider that 100a+10b+c = 99a + 9b + (a+b+c). Suppose that the sum of the digits is divisible by 9, then a+b+c=9k for some constant k. Then:

100a+10b+c = 99a + 9b +9k
100a+10b+c = 9*(11a + b +k)

So the original number is divisible by 9, as required.

Edit: oops, I misread your post as saying "prove that if the sum of the digits is divisible by _9_, so is the number." Fortunately, the method of proof for 3 is essentially identical, so you should have no difficulty deriving it from what I have written here.

Answer 2

ii)
(100a+10b+c) - (a+b+c) = 99a+9b

iii)
100a+10b+c= 99a+a+9b+b+c =
99a+9b+(a+b+c)

if the sum of digits is divisible to 3 then a+b+c = 3K
so the number can be written as

99a+9b+3K or
3(33a+3b)+3K or
3(33a+3b+K)

33a+3b+K is an integer, let be called L
so the number now can be written as 3L which is divisable to 3

a rule can not be proved by showing any number of cases in which it is true, but could be rejected by giving at least one example in which it proves false
what you should do is to prove in any logical way that it works in all conditions just like the proof given in iii) above

good luck

Answer 3

i) Because this is the Arabic-Indian system, that have made calculations easier.



iii) Let n be a natural number of form n=3m+k, k in {0, 1, 2}

If n<10 There's noting to prove.

Now, suppose it is true for any n>=10, and prove for n+3
Split n into n=10n1 + n2, n1,n2 are both positive integersl, 0<=n2<=10

Let m1,k1 be numbers such that n1 = 3*m1 + k1.
Do the same for n2, m2, k2

n = 10*(3m1 + k1) + 3m2 + k2 = 30m1 + 10k1 + 3m2 + k2

30m1 +3m2 are divisible by 3 and smaller than n, so we can assume there sum of digits s1 is divisible by 3

Now let s2 = k1 + k2.

The sum of digits of 10k1 is k1.

10k1 + k2 = 9k1 + k1 + k2,

So, the remainder of dividing 10k1+k2 by 3 is the same as of k1+k2.

and of the overall sum of the digits of (30m1 + 3m2) + 10k1 + k2