## Wednesday, 26 October 2011 See Maths Challenges (Questions)...

Answer1: When two distinct lines cut each other, they always meet at single point. These two lines can cut the circle maximum at four different points. Therefore, maximum number of intersection points is 5 i.e (d). Following figure explains this.

Total possible outcomes are 16 i.e.
HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT
The catch here is at least as many heads as tails, such possibilities are 11.
$\Rightarrow$ the probability is $\frac{11}{16}$

Here product is a prime number. Since we know prime number is divisible by self and number 1 only. So the only choice for m is 1.

Consider the worst scenario,
Number of attempts to open the first box:  9
Number of attempts to open the second box:  8
Number of attempts to open the third box:  7
and so on...

$\Rightarrow$ Total number of attempts are:

$9 + 8 + 7 + 6 + \ldots + 2 + 1 = \frac{(9)(9+1)}{2}= 45$

Out of 32 only one will be the champion, so total number of matches to be displayed will be 31.

Total things in Vats are:
$4 Mats + 8 Cats + 16 Hats + 32 Rats + 64 Bats = 124$

Let us calculate first how many numbers are there less than and equal to 999 which are divisible by 7 = $999 \div 7 = 142.714 = 142 numbers$

Numbers divisible by 7 and divisible by 11 would be divisible by 77 also.
$\Rightarrow$ Numbers divisible by 77 (<= 999) are: $999 \div 7 = 12.97 = 12$
$\Rightarrow$ Numbers divisible by 7 (<= 999)  but not by 11 = 142 - 12 =  130

Check the patterns in the number 9797797779777797777797777779...?
The 2nd 9 comes after how many 7s = 1
The 3rd 9 comes after how many 7s = 2
The 4th 9 comes after how many 7s = 3 and so on.
Therefore, the 100th 9 comes after how many 7s = 99.
$\Rightarrow$ Number of 7s appearing before 100th 9 are:
= 1 + 2 + 3 + 4 + ... + 99 = $\frac{99\times 100}{2} = 4950$
Since the number 9s appeared = 99
Total number of digits before 100th 9 = 4950 + 99 =  5049  