Arithmetic+  05: UNIZOR.COM - Math+ & Problems - Arithmetic
Channel: Zor Shekhtman on UNIZOR Education
Date: 2024-03-02

UNIZOR.COM - Creative Mind through Art of Mathematics
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Arithmetic+ 05

Problem A
Prove that a remainder of division of any prime number P by 24 is a prime number, assuming P is greater than 24.

Hint A
If P is prime and R is a remainder of division of P by 24 then P=24·n+R where R≤23.
If R is not prime, it should be divisible by a product of, at least, two prime numbers.
Also, 24=2·2·2·3.


Problem B
Prove that for any natural number N the number (10^N)+(18·N)−1 is divisible by 27.

Hint B
First, prove a theorem that any natural number n divided by 3 (or 9) has the same remainder as a sum of its digits divided by 3 (or 9).
Obviously, the given number is divisible by 9.
Then prove that the number 111...111 consisting of N unit digits with their sum equal to N plus 2·N is divisible by 3 using a theorem proved above.


Problem C
Prove that from any set of 100 natural numbers N1, N2,... ,N100 it is possible to choose such a subset with a sum of its numbers divisible by 100.

Hint C
Let Sk=Σi∈[1,k]Ni
where k=1, 2,...,100.
If there is one particular number Sk divisible by 100, the components of this sum can be chosen as a subgroup and the problem is solved.
Assume, none of our 100 sums Sk is divisible by 100.
Let Rk is a remainder of division of Sk by 100.
Note that there are 99 possible remainders from 1 to 99.