practice problems pending

Q1. 100 in base 10 needs to be converted to base 2,3,4,5,6,7,8,9.
S2.
Base 9: 100 = 81 + 18 + 1 = 121
Base 8: 100 = 64 + 4.8 + 4 = 144
Base 7: 100 = 98 + 2 = 202
Base 6: 100 = 72 + 24 + 4 = 244
Base 5: 100 = 25*4 = 400
Base 4: 100 = 64.1 + 16.2 + 4 = 1210
Base 3: 100 = 3^4.1 + 3^2.2 + 1 = 10201
Base 2: 100 = 2^6.1 + 2^5 + 2^2.1 = 1100100 

Q2.

S2.
Let's try.
Last digit is 1.
Then 5+4 = 9 but it says 2.
So it is not base 10.
Looks like base 7.
And since there is enough gap in the beginning of the first number, 2 numbers will fit there.
Answer:
23451 + 15642.


Q3. Let n be a positive integer and d be a digit such that the value of the numeral 32d ​ in base n equals 263, and the value of the numeral 324 ​ in base n equals the value of the numeral 11d1 in base six. What is n+d?
S3.

3n^2 + 2n + d = 263
3n^2 + 2n + 4 = 216 + 36 + 6d + 1

Solving for d we get d = 2
And solving for n we get n = 9 and another non integer.
Answer n + d = 11

Q4.

Using the digits 1,2,3,4,5,6,7, and 9 , form 4 two-digit prime numbers, using each digit only once. What is the sum of the 4 prime numbers?

S4.
  23

+67

+41

+59

---------

190

Q5.

Does there exist a number system where the following equalities are true simultaneously:
a) 3+4=10 and 3⋅4=15

b) 2+3=5 and  2⋅3=11?

S5.

(a)

Yes.

3 + 4 = n => n = 7

12 = 7 + 5

(b)

6 = n.1 + 1 => n = 5

No, then the first one should be 2 + 3 = 10

Q6.

State and prove the test for divisibility by

a) a divisor of the number  n−1 in the base n system (similar to the divisibility test for 3 in the decimal system):

b) the number  n+1 in base n system (similar to the divisibility test for 11);

c) a divisor of the number  n+1 in the base n system.

S6.
a)
Let d be a divisor of n-1.
=> n-1 = 0 mod d
=> n = 1 mod d
=> n^m = 1^m = 1 mod d

Now let's take any random number Z

Z = p.n^k + q.n^(k-1) ... + r
Z mod d = p + q + ... r mod d

=> if 'd' divides (n-1) then it will also divide sum of its digits.

b)

n+1 = 0 mod n+1
n = -1 mod n+1
Now it's similar to c) proof.

c)
For n +1  = 0 mod d
n = -1 mod d

Z = p.n^k + q.n^(k-1) ... + r
=>
0 = p.(-1)^k + q.(-1)^(k-1) ... r mod d
=> 
If 'd' divides n+1 then it will divide any number which when expressed in base 'n' has the alternative sum of its digits divisible by 'd'.

Q7. if p,q are primes such that p>=q>=5 then pr. th. 24 | p^2 - q^2
S7.
p+q = 2m + 1 + 2n + 1 = 2(m+n+1)
p- q =2(m-n)
p^2 - q^2 = (p+q)(p-q)
= 2(m+n+1).2.(m-n)

If m+n is even then m-n is also even => m+n+1 is odd.
If m+n is odd then m-n is also odd => m+n+1 is even.
So we got another factor of 2.

So it's divisible by 8.

Now for 3, again go back to 
(p+q)(p-q)
Both p,q are not divisible by 3 so their possible remainders mod 3 are 1,2
1,1 => p-q = 0 mod 3
2,2 => p-q = 0 mod 3
1,2 and 2,1 => p+q = 0 mod 3.
H.P.