Challenging official Maths Questions .... ? Show Working?

A1 Find the number of integers n with first digit 9 where


0 %26lt; n %26lt; 100 000





A2 Find the sum of the digits of 2 to the power of 2 * 3 to the power of 3 * 2 to the power of 4 * 5 to the power of 5 when the product is multiplied out





A3 Put in increasing order:


11 to the square root of 3


7 to the square root of 4


5 to the square root of 5


3 to the square root of 6


2 to the square root of 7





A4 A games machine is programmed to set 4 switches a, b, c, d in that order to 0 or 1 according to the following system: "a is random; if a = 1 then b = 1; c = b; if a = 1 then d = c"


Which statement is certainly true?


i) a, b, c, d can't all be 1


ii) at least 3 of a, b, c, d are equal


iii) b = d


iv) at least 2 of a, b, c, d are 1


v) none of these options (i to iv) are true

Challenging official Maths Questions .... ? Show Working?
A1) that is the sum of 10^i, where i=0 to i=4


that is


10^0+10^1+10^2+10^3+10^4=sum


the exponents are for each digit place


1=10^0 is the ones place, there is 1 n in the one digit numbers


10=10^1 is the tens place, there are ten numbers that start with 9, etc





A2)9 = use calculator!





A3) reverse the order to get


(4*7)^.5


(9*6)^.5


(25*5)^.5


(49*4)^.5


(121*3)^.5





A4) iii


a=1; d=c=b=1


therefore d=b
Reply:A1) that is the sum of 10^i, where i=0 to i=4


that is


10^0+10^1+10^2+10^3+10^4=sum


the exponents are for each digit place


1=10^0 is the ones place, there is 1 n in the one digit numbers


10=10^1 is the tens place, there are ten numbers that start with 9, etc





A2)that is 2^7200= a 52 digit number when you add them together you will get an even number





A3) reverse the order to get


(4*7)^.5


(9*6)^.5


(25*5)^.5


(49*4)^.5


(121*3)^.5





A4) iii


a=1; d=c=b=1


therefore d=b
Reply:use Visual Basic. It will help you. i already got the answer, but im not gonna tell you what it is.