Math Geniuses!! Please help me with this Problem of the Week...?

Question

Math Geniuses!! Please help me with this Problem of the Week...?
I'd be passing Trig if i get this one! So please help me out as much as you can, and i'll love you forever :)
WARNING: THIS IS SUPER HARD. NO ONE CAN FIGURE IT OUT IN MY CLASS (and i have a smart class)

Find the sum of all prime numbers between 1 and 100 that are simultaneously one greater than a multiple of 4 and 1 less than a multiple of 5.

This is confusing i have no idea what they're talking about.

Thanks guys!!! Even if you dont get it, i give you credit for trying!!!

Answer 1

do you even know what a prime number is??????? Any number whose factors are itself and the number one. So, you need to add all of the prime numbers between 1 and 100 together that are simutaneously one greater than a multiple of 4 and 1 less than a multiple of 5.

Answer 2

29+89 = 118

Prime numbers with the exception of 2 are always odd. Any number one less than any multiple of 5 ends in 4 or 9. A prime number cannot end in 4. All numbers ending in 9 between 1 and 100 are:

9 19 29 39 49 59 69 79 89 99

The only primes in this group are:

19 29 59 79 89

The only ones that are one greater than a multiple of 4 are 29 and 89:

29 + 89 = 118

Answer 3

Here is another way to do this.
Let x = 1 + 4t and x = -1 + 5u.
Then 1 + 4t = -1 + 5u
4t = -2 + 5u
t = 2 + 5u.
The last step comes from the fact that -2/4 = -1/2 = 2
on a 5 hour clock. Why?
-1/2 = 2 implies -1 = 4, and if we go 4 units clockwise
and 1 unit counterclockwise, we end up in the same place!
Now x = 1 + 4(2 + 5u) = 9 + 20u. But
the only primes less than 100 of the form 9 + 20u
are 29 and 89, so their sum is 118.
BTW. This technique is an example of
the Chinese Remainder Theorem, and
was known in China in 1200 BC!

Answer 4

I don't know how much help this will be, but it is worth noting that x^3 - y^3 can be factored as shown below. X^3 – Y^3 = (X – Y)(X^2 + XY + Y^2) We can rearrange this as X^3 – Y^3 = (X – Y)([X^2 + Y^2] + XY) From the information given we know that X - Y = 2 and X^2 + Y^2 = 8, so all we need is to find XY. Let's use X - Y = 2, and square both sides. (X - Y)^2 = 2^2 X^2 -XY -XY + Y^2 = 4, do some strategic grouping [X^2 + Y^2] - 2XY = 4, from above we know that X^2 + Y^2 = 8 8 - 2XY = 4, subtract 8 from both sides 8 - 8 - 2XY = 4 - 8, do the math -2XY = -4, divide both sides by -2 -2XY/-2 = -4/-2, do the math XY = 2 Now using the rearranged equation from above we can do some substitution X^3 – Y^3 = (X – Y)([X^2 + Y^2] + XY) = (2)(8 + 2) = (2)(10) = 20 Always remember that for these type of problems you will most likely have to do some algebraic manipulation.

Answer 5

Well, the first thing I'd do is make yourself a list of primes between 1 and 100.
(There's really not that many... Start with 2, 3, 5, 7, 11, 13, 17, 19... and you're already 20% done with the list!)
Then check each one (really, there aren't too many!) for your qualifications:
Like 2 & 3 obviously aren't one more than multiples of 4... but 5 is -- but it isn't one less than a multiple of 5...

This genius problem looks really easy to me...
It'll just take a little bit of patience.

...And then add up the ones you find when you're done (as it says in the problem).

Barjesse37

Answer 6

First list all prime numbers between 1 and 100. Then just erase all of the numbers that have a multiple of four and five (numbers that can be divided by 4, and 5) because one greater than a multiple of four is 5, and one less than a multiple of 5 is 4...i think......
then just add up all of the numbers that you didn't erase....then the sum of that should be the correct answer.....i think...

GOOD LUCK!

Answer 7

ALL PRIME NUMBERS ADDED TOGETHER WITHOUT THE ONES THAT WOULD BE A NUMBER GREATER THAN A 4 MULTIPLE OR LESS THAN A FIVE MULTIPLE

Answer 8

29 and 89
29(7x4=28)one higher =29
89(18x5=90)one less=89
29+89=118

Answer 9

i think the answer is 118, i tried it and thats what i got,though i cant really explain how to do it

Answer 10

4x2=8
5x2=10
so.. 9 is one!
I think i'm doing it wrong though..sorry
not enough intelligence in me.