Maths Quiz (Remainders)

(No calculator!) Some modular arithmetic fun for the number theory maths lovers. I don't expect it to be for everyone, but hopefully there are a couple of people who will enjoy this.

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Last updated: May 1, 2023

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First submitted	December 22, 2022
Times taken	7
Average score	50.0%
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1. The largest non-square remainder of a square number divided by 11

3 ✓

5 ✓

7 ✓

9 ✓

10 ✓

Consider up to 11^2, and greater than 3^2 (<11). 16,25,36,49,64,81,100-> 5,3,3,5,9,4,1

2. The final digit of 7^2023

1 ✓

3 ✓

7 ✓

9 ✓

7^1->7, 7^2->9, 7^3->3, 7^4->1, ... and we repeat the pattern, as 2023 in form 4k+3, it is 3.

3. Remainder when 20 divides 2023^2022

1 ✓

3 ✓

7 ✓

9 ✓

20 divides 2020, so we only need to worry about the 3. 3^1->3, 3^2->9, 3^3->7, 3^4->1 and we repeat. 2022 in form 4k+2 so it is 9.

4. Remainder when x^2 + 2023 is divided by x - 1, for sufficiently large x

0 ✓

2022 ✓

2023 ✓

2024 ✓

Undefined ✓

x^2 - 1 + 2024 = (x+1)(x-1) + 2024, therefore 2024/(x-1) is the part that gives remainder.

5. The remainder when working out the mean value of the factors of 10,000.

1 ✓

4 ✓

9 ✓

11 ✓

36 ✓

10 = 2^4 5^4, number of factors = 5*5 = 25. Sum of factors = (1+2+4+8+16)(1+5+25+125+625) the right bracket = 6 mod 25, as all 5^n n>2 divisible by 25. Left bracket: 16 + 8 + 1 = 25, so we're left with 2 + 4 = 6. 6* 6 = 36. 36 > 25-> 36-25 = 11.

6. The remainder of the value when dividing n^3 + 1 by n+1, in terms of n (n is integer)

0 ✓

n^2-n+1 ✓

n-1 ✓

n ✓

n+1 ✓

n^3 +1 = (n+1)(n^2-n+1) therefore for integer n n+1 divides n^3 + 1 without remainder.

7. The type of sequence the remainders form for n>3 when you divide (n+1)^2 by n^2

Arithmetic ✓

Geometric ✓

Recursive ✓

Quadratic ✓

No meaningful pattern ✓

(n^2 + 2n + 1) mod n^2 = 2n + 1 mod n^2. This is arithmetic.

8. The final digit of a 9-fingered civilisation writing the number 2023.

1 ✓

3 ✓

5 ✓

7 ✓

9 ✓

2+0+2+3 = 7 therefore it is 7.

9. The largest power of 3 that divides 1234567891011...4950

3^0 ✓

3^1 ✓

3^2 ✓

3^3 ✓

3^50 ✓

Sum of 1 to 50 = 25*51. 25 is not multiple of 3, 51 = 3*17. Therefore 3*5^2*17, meaning it is 3^1.

10. What is the last non-zero digit of 20! ?

2 ✓

4 ✓

6 ✓

8 ✓

20! = 1*2*3*4*5... get rid of all groupings of 10's -> 3*2*6*7*8*9*11*12*13*14*3*16*17*18*19*2 -> Looking at only last digits of products: 6*2*2*2*2*8*6*8->2*4*6*8->8*8->4

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