maths whiz?

Question

1) show that the remainder when 30^99+61^100 is divided by 31 is 0.
2)find the remainder when the sum 1^2 +2^2...+100^2 is divieded by 4.

3) Show that 8|5^2n +7

4) Among the rules of divisibility. Perhaps the most famous one is the "casting-out 9s"rule. A natrual number n is vidisible by 9 if and only if sum of its digits is divisible by 9. Prove it.

Answer 1

Work directly with the remainders :

30 mod 31 = -1
thus 30^99 = -1^99 = -1 ( all mod 31 )
61 = -1 mod 31
thus 61^100 = -1^100 = 1 mod 31

from the above observations follows that 30^99+61^100 is divided by 31 is 0.

2)
1^2 +2^2...+100^2 = x mod 4
1 mod 4 = 1
2^2 mod 4 = 0
3^2 mod 4 = -1
4^2 mod 4 = 0
5^2 = 1 mod4 ( here sequence repeats
thus x = 0.

3)
show that 8|5^2n +7
for n = 1 : 25 + 7 = 32 is divisible by 8. thus ok.
now for n>1 :
5^2n + 7 = (5^2(N-1) + 7)*5^2 - 7*24
now the first part : (5^2(N-1) + 7)*5^2 is divisible by induction hypothesis, and 7*24 is divisiv-ble by 8
qed.

4) the 9-rule relies on the fact that :

(An . x^n + An-1 . x^n-1 + .... +A0 ) mod P is equal to the sum of the mods of the individiual terms mod P
( this I used in 2) and 1) )

1 mod 9 = 1
10 mod 9 = 1
100 mod 9 = 1, etc
thus for instance 45 mod 9 = 4 mod 9 + 5 mod 9 = 9 mod 9 = 0 mod 9

Answer 2

1) 30^99 + 61^100 =(61-31)^99 + 61^100 = 61^99 + terms containing 31 + 61^100.
=61^99 + 61^100 + terms containing 31.
=61^99(1 + 61) +terms containing 31.
=31 x 2 x 61^99 + terms containing 31 which is divisible by 31

2)1^2 +2^2 +...+100^2 = (!00 x 101 x 201) /6 =50 x 101 x 67
= 50 x 100 x 67 + 50 x 67.
Enough to find the remainder when 50 x 67 is divided by 4 leaving the hundreds.
Enough to find the remainder when 50 x 7 is divided by 4 Enough to find the remainder when 50 x 1 is divided by 4 4. The remainder is 2

3) We prove this result by induction.
The result is true for n = 1, as 8 divides 32.
Assume the result for n = m. Thus 8 I 5^2m + 7.
Now 5^2(m+1) + 7 = (5^2m)(5^2) + 7 = 5^2m x 25 + 7= (5^2m +7) + 5^2m x 24 which is divisible by 8.
The result is true for all n.

4)For the sake of simplicity consider a 4 digit number.
abcd = ax1000 + bx100 + cx10 +d =a(999+1) + b(99+1) + c(9+1) +d = 999a +99b + 9c + (a+b+c+d) which is divisible by 9 if a+b+c+d is divisible by 9.

Answer 3

For 1), I tried without using mod rules:

30^99 + 61^100
= 30^99 + (30+31)^100
= 30^99 + (30^100 + 31(the remaining expansion of the 2nd term above))
= 30^99 + 30 x 30^99 + 31(...)
= (1 + 30) x 30^99 + 31(...)
= 31 (...)
thus remainder is 0.

2) 1^2 + 2^2 + 3^2 + ... + 100^2
= 1^2 + 2^2 + (2+1)^2 + 4^2 + ... + (98+1)^2 + 100^2
= 1^2 + 2^2 + (2^2 + 2x2x1 + 1^2) + 4^2 + ... + (98^2 + 2x98x1 + 1^2) + 100^2
= 1 + 4 + (4 + 4 + 1) + 4x2^2 + ... + (4x49^2 + 4x1x49x1 + 1) + 4x50^2
= 1 + 1 + ... 1 (50 of them) + 4 x (1+1+1+2^2+...+49^2+49+50^2)
= 50 + 4(...)
thus remainder is 2.

3) 5^(2n) = 5^2^n = 25^n
For n = 1, it is 025, for n=2,3,4,... its last 3 digits are 625.
Thus last 3 digits of 5^(2n)+7 are either 032 or 632.
My observation, and should be provable, that if 3rd last digit is even and last 2 digits form a number divisible by 4, the whole number is divisible by 8.
So, 5^(2n)+7 is divisible by 8.

Answer 4

1) 30^99 = (31-1)^99

When you expand this binomial, all terms will have 31 in it but the last one which is -1^99
so the remainder is -1

61^100 = (62-1)^100
Expand it and the only term without 62 (31 X 2) is -1^100
so remainder is +1

Add both the terms and the remainder is -1 + 1 = 0

2) sum of n square terms is n(n+1)(2n+1)/6

Put n = 100, you have 100 X 101 X 201

This when divided by 24 (6X4)...gives the remainder 1
so your answer is 1

3) 5^2n + 7 = (8-3)^2n + 8 - 1
Expand and you have all terms as multiple of 8 but,

(-3)^2n -1 = 3^2n -1
This is same as 9^n - 1
or (8+1)^n - 1

Expand and you get

remainder as 1^n - 1 = 0
So, 5^2n + 7 is always divisible by 8

4) Lets a there is a number xyz...dcba any number of digits.
Then this number is equal to

a + 10b + 100c + 1000d +.....
This is same as a + b + 9b + c + 99c + d + 999d + ...

all terms are div by 9 except... a + b+ c + d + ... x + y + z

So, for this number to be div by 9, sum of its digits must be a factor of 9.

Answer 5

1) Hint: 30=-1 and 61=-1 modulo 31.

2) Hint: Squares are either 0 or 1 modulo 4. How many 1's do you get?

3) Hint: 5^2=25=1 modulo 8.

4) Hint: 10=1 modulo 9.

You will have to think more about this stuff if you plan to get through that number theory class.

Answer 6

two times the dimensions further to 2 times the width is the fringe. L = w+5 P = 2l+2w plug interior the 1st equation into the 2nd (plug interior the L): P = 2(w+5) + 2w replace P with 54cm and distribute the two: fifty 4 = 2w + 2w + 10 Isolate w via including like words and subtracting 10 from the two factors: fifty 4 - 10 = 2w+2w +10 -10 40 4 = 4w Divide via 4 on the two factors: 4/4*w = 40 4/4 w = 11 hence L = 11+5 = sixteen... Dimensions: 16cm x 11cm

Answer 7

4)casting out of nine
613 =6+1+3=[10]-9={ 1 }
*85 =8+5= [13]-9={ 4 }
{ 1 }*{ 4 }=[ 4 ]

631*85=52105
52105 =5+2+1+0+5=[13]-9=[ 4 ]

Answer 8

What are you in? Grade School!
There is nothing remotely challenging there (well, except for working out what you're talking about with the third one.)

Answer 9

huh