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Junior | Next Rank: 30 Posts
Posts: 29
Joined: Tue Nov 24, 2009 12:48 am

by Pooja Bhula » Wed Nov 25, 2009 1:47 am
heshamelaziry wrote:
2010gmat wrote:great ques hesham...

if the question says at least 3 heads in 5 tosses...

then??

will it be

5c0 + 5c1 + 5c2 + 5c3)/2^5 ???i.e 30/32
I don't know 2010gmat. I am poor in this subject. I just memorize question patterns, unfortunately. I never took a class that covers all this stuff.

It surely cant be 5C0 becoz at least means 3 or more than 3...but even I want to know the answer to this, can someone please help? i think it wud be (1/2)^3 + (1/2)^4 + (1/2)^5 = 7/32... this is the 'OR' or addition principle.. can someone tell me if this is surely correct?
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Junior | Next Rank: 30 Posts
Posts: 29
Joined: Tue Nov 24, 2009 12:48 am

by Pooja Bhula » Wed Nov 25, 2009 2:31 am
heshamelaziry wrote:
Pooja Bhula wrote:Hi Hesham,
Im gonna help you with the sum to find a^36. Now see we are only concerned with the units digit here.. so lets see which square(s) gives us the units digit 9, n which cube(s) give us 3 in the units digit. So here go the squares - 2 = 4, 3 = 9, 4 = 16, 5 = 25, 6 = 36, 7 = 49.thus only 3 and 7 end in units digit 9, when they are squared. Therefore, lets see the cubes of 3 and 7 to decide which no. it is. 3^3 = 27, so this does not work coz we need 3 in the units place. Now take 7, 7^3 = 49 x 7 = 343. So bingo.. 7 is our no.

Now the next step is to find a pattern.... for 7^n... 7^1 = 7, 7^2 = 49, 7^3 = 343, 7 ^4 = here the units digit ends in 1 as 7 x 3 = 21, next multiply the units digit by 7..so 7^5 = units digit is 7. Thus every 4th one ends in 1, and 36 being a multiple of 4 also ends in 1, or has 1 in the units place.
Is the answer b or d in your opinion ?
I think it is b...a alone cannot give you a definate ans as 3, 7 both have 9 in units digit wen squared...but b tells u d cube n only 7 cube ends wid 3 in units place...
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