Number theory: Numbers with different bases, Cyclicity in multiplication

Scale notation: (Digital number system)

Decimal ⇒ base 10 ✔
→ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

123 = 1 × 10² + 2 × 10¹ + 3 × 10⁰

(2437)₁₀ = 2 × 10³ + 4 × 10² + 3 × 10¹ + 7 × 10⁰

(3412)₅ = 3 × 5³ + 4 × 5² + 1 × 5¹ + 2 × 5⁰

(123)₍b₎ = 1 × b² + 2 × b¹ + 3 × b⁰

Ex1:

Q. On certain number system (given that base is not negative)

    363 + 1056 = 1452

Find the value of 654 − 456 in same number system

3 × b² + 6 × b + 3 + 1 × b³ + 0 + 5 × b + 6  

= 1 × b³ + 4 × b² + 5 × b + 2  


→ Combine like terms:  

3b² + 6b + 3 + b³ + 5b + 6 = b³ + 4b² + 5b + 2  


→ Cancel matching terms:  

b² - 6b - 7 = 0  


Solve:  

b² - 6b - 7 = 0  

b² - 7b + b - 7 = 0  

b(b - 7) + 1(b - 7) = 0  

(b - 7)(b + 1) = 0  


b = 7  ✅  

b ≠ -1 ❌

Now convert the given expression to decimal:
654 - 456 = 6*7^2 + 5*7 + 4 - (4*7^2 + 5*7 + 6) = 2*7^2 - 2 = 96 base 10
To convert a base 10 number to base 7(or any other base), here is a method:
1. Divide the number N by the new base, note the quotient(Q) and remainder(R).
2. Replace N = Q and repeat.
3. Write down all the remainders in sequence and at the end write the last quotient, that's the answer.

Let's convert 96 base 10 to a base 7 number.
96 % 7  = Q:13 R: 5
13 % 7 = Q:1 R: 6
Answer: 165
Double check: 1*49 + 6*7 + 5 = 96 base 10. Correct.

Ex2:

Find x.

(3412)₅ = (x)₁₀

Solution:
In the last question we saw that for conversion to base 10 from non base 10 we simply write down the powers.
3*5^3 + 4*5^2 + 1*5 + 2 = 375 + 100 + 5 + 2 = 482 base 10

Ex3:

If  (25)ₙ × (31)ₙ = (1015)ₙ
Then find the value of

(13)ₙ × (52)ₙ, given n > 0
Solution:
(2n + 5)(3n + 1) = n^3 + n + 5
=> 6n^2 + 17n  = n^3 + n 
=> n^3 - 6n^2 - 16n  = 0
Base can't be 0 so n^2 - 6n - 16 = 0 => n = 8,-2
But n > 0 => n = 8
So given expression becomes:
11*42 = 462 base 10

Ex4:

Q. How many two-digit numbers satisfy the property:
When the number is added to the number obtained by reversing its digits, the sum is 132?

Solution:
Let the number be ab.
10a + b + 10b + a = 11(a+b) = 132 => a+b = 12
Possible pairs of (a,b):
(3,9)
(4,8)
(5,7)
(6,6)
and reverse so total 7 numbers.

Ex5.

Q.
(25)x=(85)10(25)_x = (85)_{10}

Find x.

Solution:
2x + 5 = 85 => x = 40

Q.

49−x2−25−x2=3\sqrt{49 - x^2} - \sqrt{25 - x^2} = 3

What is:

49−x2+25−x2= ?\sqrt{49 - x^2} + \sqrt{25 - x^2} =\ ?

Solution:

Let:

49−x2+25−x2=K\sqrt{49 - x^2} + \sqrt{25 - x^2} = K

Multiply both to get:
49 - x^2 - (25 - x^2) = 3k => 3k = 24 => k = 8. Answer.

New Topic: Cyclicity of 1,2,3,4,5,6,7,8,9,0:
Let's take digit 4.
4^1 = 4
4^2 = 16
4^3 = 64
4^4 = 256
So the unit digit alternates between 4,6.
Similarly for 9, the unit digit alternates between 9,1.

For 0,1,5,6:
Unit digit always remains same no matter how many times you multiply the digit.

For 2,3,7,8 the period for repetition is 4:
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
....
3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 3
....
7^1 = 7
7^2 = 49
7^3 = 343
7^4 = 2401
7^5 = 16807
....
8^1 = 8
8^2 = 64
8^3 = 512
8^4 = 4096
8^5 = 32768

So if we need to find last few digits for a large number we can use these properties.

Ex1:
Find unit digit of:

Q.

(64354)1237+(67354)1236(64354)^{1237} + (67354)^{1236}

Solution:
4 has cyclicity of 2 so the first number's unit digit is 4 and for the second one it's 6.
Adding 4 and 6 we get the unit digit  = 0 = Answer.

Ex2:
Find unit digit of:

Q.

(127)39+(127)38(127)^{39} + (127)^{38}

Solution:
7 has cyclicity of 4 so effectively it's 7^3 + 7^2 = 3 + 9 = 2 is the unit digit.

Ex3:
Find unit digit of:

Q.

11+22+33+44+⋯+10101^1 + 2^2 + 3^3 + 4^4 + \cdots + 10^{10}

Solution:
1^1 = 1
2^2 = 4
3^3 = 7
4^4 = 6
5^5 = 5
6^6 = 6
7^7 = 7^3 = 3
8^8 = 8^4 = 6
9^9 = 9^1 = 9
10^10 = 0
Add all to get: 47. Answer = 7.

Ex4.
Find unit digit of:
(1!)^(1!) + (2!)^(2!) + (3!)^(3!) .... (1000!)^(1000!)

Solution:
The trick here is that starting 5!, each number is a multiple of 10.
So no matter which power you take the unit digit will be 0.
So we just focus on first 4 numbers.
1
4
6
24^24 = 4^4 = 6
Answer: 1 + 4 + 6 + 6 = 17 = 7

Ex5.
Find unit digit of 222^888 + 888^222.
Solution:
6 + 4 = 0

Ex6.
Find unit digit of:
12^(1! + 2! + 3! ... 10!)
+ 13^(1! + 2! + 3! ... 10!)
+ 14^(1! + 2! + 3! ... 10!)
.....
+ 20^(1! + 2! + 3! ... 10!)

Solution:
Remember that maximum cyclicity of any digit is 4. So we can discard (in the powers) all the multiples of 4 and just focus on the rest.
Starting 4!, all numbers are multiples of 4.
Remaining power = 1 + 2 + 6 = 9
Again remove 8 and only 1 remains.
So just add the unit digits of the given numbers:
2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 0 = 9*10/2 - 1 = 44 = 4 = Answer.

Ex7.
Find the right-most non-zero digit of 10!.
Solution:
10! = 1*2*3*4*5*6*7*8*9*10
It has 2 zeroes made from 2,5,10.
So the last 2 digits are 00.
We have to multiply the rest and find the unit digit.
3*4*6*7*8*9 = 8 = Answer.

Ex8.
Find last three digits of:
5^2 * 2^100 * 33^99 * 77^88 * 88^66

Solution:
Again last 2 digits are 00 since 25*4 = 100.
Remaining:
2^98 * 33^99 * 77^88 * 88^66
= 4 * 7 * 1 * 4 = 2
So last 3 digits = 200.

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