Tutor MathsGuru: Ask me for your burning Maths questions!

[Herbie]

can help with the qn below??

a stairs with 10 steps. U can either take one step or 2 step at a time. How many ways can u go up the stairs?

Tq

[MathIzzzFun]

Herbie:

can help with the qn below??

a stairs with 10 steps. U can either take one step or 2 step at a time. How many ways can u go up the stairs?

Tq

Hi

math olympiad question?

I think you can find solutions to similar type question in the math olympiad thread. You can solve this using the Fibonacci numbers.

Nx = number of ways to go up x steps

Number of ways to go up a 10 steps, N10
= Number of ways to go up 9 steps and then do a 1-step
+ Number of way to go up 8 steps and then do a 2-step
=N9+N8

working backwards, we have N9=N8+N7, N8=N7+N6... N3=N2+N1
N1=1
N2=2
N3=1+2=3
N4=2+3=5
N5=3+5=8
N6=5+8=13
N7=8+13=21
N8=13+21=34
N9=21+34=55
N10=34+55= 89

there are 89 ways to walk up 10 steps using 1 step or 2 steps.

cheers.

[Herbie]

thanks mathizzzfun.

[MathIzzzFun]

Herbie:

thanks mathizzzfun.

u r welcome

cheers.

[Essential]

Math Guru, pls help me with this question

Renjie has three ropes with length of 144,168 cm and 252cm. He wants to cut the three ropes into shorter pieces of equal length with no remainders a) what is the greatest possible length of each of the shorter pieces? b) how many of the shorter pieces of equal length can he get?

[MathIzzzFun]

Essential:

Math Guru, pls help me with this question

Renjie has three ropes with length of 144,168 cm and 252cm. He wants to cut the three ropes into shorter pieces of equal length with no remainders a) what is the greatest possible length of each of the shorter pieces? b) how many of the shorter pieces of equal length can he get?

Hi

not math guru..mathizzzfun here..

this question is about HCF = highest common factor.

The standard way is the list out the factors for each number and then pick up all the common factors.

However, for these 3 numbers, we could tak a little short cut..
first, we see that all the 3 numbers are divisible by 3 (test divisibility by 3 --> sum of digits divisible by 3) and all numbers are also divisible by 4 (test divisibility by 4 --> last 2 digits divisible by 4)

So, 4x3 =12 is a factor for all the numbers...
144 = 12 x 12 = 12 x 2 x 2 x 3
168 = 12 x 14 = 12 x 2 x 7
252 = 12 x 21 = 12 x 3 x 7
There are no more common factor, so 12 is the HCF --> Longest length for each piece = 12 cm
Number of 12cm pieces = 12 + 14 + 21 = 47

cheers.

[Essential]

Thank u Marhizzzfun

[MathIzzzFun]

Essential:

Thank u Marhizzzfun

u r welcome

cheers.

[kwcllf]

Hi please help with the following question.


Roy, Sam and Ted had a sum of money. Roy had 4/5 of Sam’s money. Sam had 60% of Ted’s. After Roy gave $12 to Sam, he had 5/7 of what Sam had. How much more money did Ted have than Sam in the end?

Thanks:-)

[Essential]

Hi, Need help on this question.

1) A rectangular cardboard measures 33 cm by 24 cm. What is the maximum number of rectangular pieces that can be cut from the cardboard if each piece of rectangle measures t6 cm by 4cm? Please help me confirm the answer is 28 , 32 , 30 or 34
Thanks.