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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 | 12 |
| Average score | 40.0% | Report this quiz | Report |
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1. The largest non-square remainder of a square number divided by 11
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
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5
7
9
10
2. The final digit of 7^2023
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.
1
3
7
9
3. Remainder when 20 divides 2023^2022
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.
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7
9
4. Remainder when x^2 + 2023 is divided by x - 1, for sufficiently large x
x^2 - 1 + 2024 = (x+1)(x-1) + 2024, therefore 2024/(x-1) is the part that gives remainder.
0
2022
2023
2024
Undefined
5. The remainder when working out the mean value of the factors of 10,000.
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.
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4
9
11
36
6. The remainder of the value when dividing n^3 + 1 by n+1, in terms of n (n is integer)
n^3 +1 = (n+1)(n^2-n+1) therefore for integer n n+1 divides n^3 + 1 without remainder.
0
n^2-n+1
n-1
n
n+1
7. The type of sequence the remainders form for n>3 when you divide (n+1)^2 by n^2
(n^2 + 2n + 1) mod n^2 = 2n + 1 mod n^2. This is arithmetic.
Arithmetic
Geometric
Recursive
Quadratic
No meaningful pattern
8. The final digit of a 9-fingered civilisation writing the number 2023.
2+0+2+3 = 7 therefore it is 7.
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5
7
9
9. The largest power of 3 that divides 1234567891011...4950
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.
3^0
3^1
3^2
3^3
3^50
10. What is the last non-zero digit of 20! ?
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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6
8
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