Arithmetic+ 04: UNIZOR.COM - Math+ &Problems - Arithmetic

Notes to a video lecture on http://www.unizor.com

Arithmetic+ 04

Problem A
Consider two numbers:
X=11111111
(digit 1 repeats 8 times) and
Y=11111...11111
(digit 1 repeats 100 times).
What is their greatest common factor?

Hint A
Both numbers can be represented as a sum
10⁰+10¹+...+10^k
where k=7 for number X and k=99 for number Y.
From this it's easy to see that
Y = X·n + 1111·10⁹⁶
where n is some natural number.

Answer A
The greatest common factor of numbers X and Y is 1111.


Problem B
Prove that a remainder of the division of some natural number by 9 is the same as a remainder of the division by 9 of the sum of this number's digits.

Example B
Take, for instance, number 2024 and divide it by 9.
2024÷9 = 224 (8)
(remainder is 8)
Now with the sum this number's digits:
(2+0+2+4)÷9 = 8÷9 = 0(8)
(the same remainder 8)

The rule of divisibility by 9
Consequently, if the sum of digits of a natural number is divisible by 9, then the number itself is divisible by 9.

Hint B
Represent a number as a sum of its digits, each multiplied by a factor 10ⁿ, where n=0 for the right most digit, n=1 for the next digit to the left etc.


Problem C
Prove that if the sum of digits of some natural number N is the same as the sum of digits of the number 2·N then number N is divisible by 9.

Example C
For number 279 the sum of digits is 2+7+9=18.
Double this number:
2·279 = 558
Sum of digits of 558 is also 18.
The number 279 is indeed divisible by 9:
279÷9 = 31 (0)
(remainder is 0)

Hint C
Use the result of the Problem B above and consider the difference between 2·N and N.

Proof C
N÷9 = n (r)
(here n is a quotient and r is a remainder with the value from 0 to 8)
Therefore, N = 9·n + r
Analogously,
(2·N)÷9 = m (r)
(same remainder r)
2·N = 9·m + r
Subtract formula for N from formula for 2·N
2·N − N = N =
= (9·m + r) − (9·n + r) =
= 9·(m−n)
Hence
N = 9·(m−n)
that is, N is divisible by 9.

Note
The same result (divisibility by 9) can be obtained under weaker conditions.
The theorem can be stated as follows.
Prove that if the sum of digits of some natural number N divided by 9 gives the same remainder as the sum of digits of the number 2·N divided by 9, then number N is divisible by 9.
Obviously, that remainder in both cases will be zero. No other equal remainder can be encountered for sums of digits of N and 2·N divided by 9.

Example CC
For number 144 the sum of digits is 1+4+4=9.
Double this number:
2·144 = 288
Sum of digits of 288 is 18.
Sums of digits are different (though, both are divisible by 9), but the number 144 is indeed divisible by 9:
144÷9 = 16 (0)
(remainder is 0)
You can easily prove it or watch the proof in the lecture.