Math Twisters?

Question

1) Find the sum of the digits of the first 200 even natural numbers.

2) If four fair dice are rolled, determine the probability that the sum is 10.

Answer 1

1. The first 200 numbers are 2, 4, 6,8 ..... 400.

Sum the digits place by place. The last place will cycle through 2, 4, 6, 8, 0 over and over. Each cycle is 5 numbers long so there are 200/5 = 40 cycles and each cycle sums to 2 + 4 + 6 + 8 + 0 = 20. So the last digit sums to 20*40 = 800.

The next digit gives you 5 zeroes, then 5 1s (in the teens), 5 2s (in the 20s) etc. Of course, it repeats again each 100. The sum of the digits from 0 to 9 is 45, there are 5 of each digit in each of teh 4 hundreds so this digit sums to 45*5*4 = 720. For what it is worth, it is exactly 90% of the previous sum and there's a good reason for that.

On to the 100s digit. It starts with 49 zeros, then 50 1s, 50 2s, 50 3s and 1 4. The 49 zeroes are because we are not counting the number zero, we start at 2. This gives a sum of 49*1 + 50*2 + 50*3 + 4 = 303.

Adding it together gives: 800 + 720 + 303 = 1823

2. Four dice can land in 6x6x6x6 = 1296 different ways. You could just count how many of them add to 10. But that's a lot. Instead break it down into 2 sets of 2 dice each. The number of ways two dice can form different numbers is easy to figure. The numbers below shown how many ways 2 dice can form the numbers 2 to 12:

2 - 1
3 - 2
4 - 3
5 - 4
6 - 5
7 - 6
8 - 5
9 - 4
10 - 3
11 - 2
12 - 1

Notice the pattern there, Now just find all the combinations that add to ten. Forexample, there are 5 ways to make a 6 and 3 ways to make a 4 so there are 5*3 = 15 ways to roll a10 by rolling a 6 with the first 2 dice and a 4 with the second two. Tabulating all the combinations that add to 10:

2/8 is 1*5 = 5
3/7 is 2*6 = 12
4/6 is 3*5 = 15
5/5 is 4*4 = 16
6/4 is 5*3 = 15
7/3 is 6*2 = 12
8/2 is 5*1 = 5

Adding them up, 80 of the 1296 ways total 10 for a probability of 0.062.

Answer 2

Sn=2+4+6+8+...............200th term
Sn makes an arithmetic sequence
which has a formula as
Sn= n(2a1+(n-1)d)/2
where n is no. of terms
a1= first term
d = difference between two terms
Sn=200(2*2+(199)2)/2=40200 ans


the next questions is too lengthy
though the trick is just find no. of ways making sum of 10
Hope that helped you.

Answer 3

FOR THE SECOND ONE A TRICK I USE IN BOARD GAMES IS TO KEEP THE DICE ON A LOW NUMBER FACING UP THAT WAY A GOOD PERCENTAGE OF THE TIME I GET A HIGH NUMBER AND VICE VERSA THIS MESSES WITH THE NATURAL PROBABILITY OF OF GETTING 10 BECAUSE IF I DID NOT LOOK AT THE DICE TO SEE WHAT NUMBER IT IS ON THEN I COULD NOT FORCE A TEN BEEN A LONG TIME SINCE I LOST A GAME OF MONOPOLY