5 More Number Theory Problems


By Yiingleong Chin on Wednesday, July 18, 2001 - 09:04 pm:

Hi!
I have got a few problems here:

1. What is the remainder when 13 +23 +33 +...+1003 is divided by 7?

2. What is the number of strictly positive roots of the equation sin x= x/200?
3. For what values of x is x2 < |2x-8|?

4. If f(x) =px7 +qx3 +rx-4 and f(-7)=3, what is the value of f(7)?

5. The x-coordinate of the foot of the perpendicular from the point (1,9) to the line y=x is ?

Can anyone show me how to solve these problems?
Thanks a lot!


By David Loeffler on Thursday, July 19, 2001 - 01:21 pm:

I'll just give you some hints for now.

1. For this one you will need some modular arithmetic. Have you come across this before?

2. Draw a picture! (Consider simpler cases first like sin x = x/10.)

3. Find all x for which x2 < 2x-8, and all x for which x2 > -2x+8. Then consider which x will satisfy both at once.

4. Let g(x)=f(x)+4. Then what is g(-x) in terms of g(x)?

5. You can do this all sorts of ways; try doing a diagram and applying some trigonometry. Alternatively, if you know about the dot products of vectors, try using these.

Have fun!


David


[Editor: See this site for details of modulo arithmetic for (1).]
By Arun Iyer on Thursday, July 19, 2001 - 07:13 pm:

for 1,
do you know the summation formula of

r=1 100 r3
if you do then find the value of the sum using that formula and find the remainder manually dividing the whole thing.

love arun
By Olof Sisask on Thursday, July 19, 2001 - 07:32 pm:

For (1), another way is to notice that (a + b) divides (a3 + b3 ). You can therefore group 13 and 63 (which is divisible by 1 + 6 = 7), 23 and 53 (divisible by 2 + 5 = 7), 33 and 43 , and leave 73 on it's own. Since each of these little 'groups' are multiples of 7, you know when you add them together, you'll get a multiple of 7, so the remainder from these terms upon division by 7 is 0. You can continue this up to 983 , but you have to consider 993 and 1003 seperately. Then, 99 = 7x14 + 1, therefore 993 = (7x14 + 1)3 = (multiples of 7) + 1, so this term leaves 1 upon division by 1. You can do the same for 1003 ( = [7x14 + 2]3 ).

Regards,
Olof


By Arun Iyer on Thursday, July 19, 2001 - 07:36 pm:

neat proof OLOF.

love arun