A number theory problem by Salmaan Suliman

Modulus is the remainder when dividing the dividend by the divisor.

Consider the equation:

A B \frac{A}{B} = Q remainder R

  • A is the dividend

  • B is the divisor

  • Q is the quotient

    • R is the remainder

For instance,

14/3 = 4 remainder 1

Therefore, 14 mod 3 = 1

Further Explanation: Khan Academy-Modular Arithmetic

Now consider the number pattern:

1 mod 10 = 1

2 mod 10 = 2

3 mod 10 = 3

...

29 mod 10 = 9

30 mod 10 = 0

From this we can see that:

let x= the last digit of n

n mod 10 = x (where x E {1;2;3;4;5;6;7;8;9;0} )

QUESTION:

T1 : 1 mod 10 = 1

T2: (1+ (T2-1)) mod 10 = (1+T1) mod 10 = 1+1 mod 10 = 2 mod 10 = 2

T3: (1+ (T3-1) + (T3-2) ) mod 10 = (1+2+1) mod 10 = 4 mod 10 = 4

T4: (1+ (T4-1) + (T4-2) + (T4-3)) mod 10 = (1+4+2+1) mod 10 = 8 mod 10 = 8

T5: (1+ (T5-1) + (T5-2) + (T5-3) + (T5-4)) mod 10 = (1+8+4+2+1) mod 10 = 16 mod 10 = 6

T6: (1+ (T6-1) + (T6-2) + (T6-3) + (T6-4) + (T6-5)) mod 10 = (1+6+8+4+2+1) mod 10 = 16 mod 10 = 22

T7: (1+(T7-1)+(T7-2)+(T7-3)+(T7-4)+(T7-5)+(T7-6)) mod10= (2+22) mod 10= 24

If we continue this pattern, we get:

T8 : 28

T9: 36

T10: 42

T11: 44

T12: 48

T13: 56

Question: Write an equation to solve for the value of Tn .

Unsolved Tn= n(n^{2} mod 10) - n/2 Tn= (Tn-1)^{2} mod 10 + n Tn = n^{2} mod 10

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