Questions

[nick3194]

1/3 + 1/6 + 1/10 + 1/15 + ... + 1/55

7^9 / 11 has a remainder of ______ (I'm assuming this involves modulus, which I never fully grasped)/

6^5 / 4 has a remainder of _______

A triangle has sides 7, 11, and k. How many integral values of k will form a triangle?

[kd9485]

This is the sum of the inverse of triangle numbers and 55 is the 10th triangle number and can be expressed as 2n/(n+1) where n is the nth triangle number.
Plugging 10 gives you 20/11 or 1 9/11. Because you didn't add the first inverse triangle number (1), subtract 1 from your answer to get 9/11.

For the next two problems, can explain the modulus stuff. Unfortunately I'm not very good at this either so I can't explain these. Sorry! I'm sure someone else could though.

There's a rule that says the sum of any two sides of a triangle is always greater than the third. So the longest possible length is 7+11>x, or 17 and the shortest is 11-7<x or 5. So that's 13 possibilities.

[michel470]

1/3 + 1/6 + 1/10 + 1/15 + ... + 1/55 Sum of reciprocated triangular numbers. Find out which triangular number the last number is. (In this case, it is 10) the answer is (n-1)/(n+1) In this case, 9/11

6^5 / 4 has a remainder of _______
To find out if a number is divisible by four, look at its last two digits. ALSO, something important to know is Euler's little theorem(I think that's what it's called ha). 6^5 divided by 4, insofar as remainders are concerned, is essentially the same thing as 2^5/4 (Find the remainder of six into four and take that to the fifth instead) 2^5=32, which has a remainder of zero.

7^9 / 11 has a remainder of ______ if you know your cubes, you'd know that 7^9 could very well be 343^3. Use Euler's little theorem, and you find that the remainder of 343 divided by 11 is 2, so it all becomes 2^3. 2^3 is 8, so the remainder is 8.

[nsguy1350]

I found a very quick trick to triangular reciprocals: if the first term is 1/3, put the second to last denominator over the last. 45/55 = 9/11.