here I will put different little exercices I find.

6th grade problem: find all integral numbers which after substraction to 9999 give the same number without the last digit. for example: abcde - 9999 = abcd damn, it must be easy

The bridge Three people (A, B, and C) need to cross a bridge. A can cross the bridge in 10 minutes, B can cross in 5 minutes, and C can cross in 2 minutes. There is also a bicycle available and any person can cross the bridge in 1 minute with the bicycle. What is the shortest time that all men can get across the bridge? Each man travels at their own constant rate. I didn't solve this problem.

Hint: if you know that d = v * t then everything is easy :|

The bicycle can be left at any place on the bridge and can be used in both directions on the bridge.

The steps are: B and C walk by foot and A take the bicycle. A at one place (y) can leave the bicycle and finish the way by foot. C can take the bicycle, drives it back to a point (x), leaves it and continues to walk by foot. B takes the bicycle and drives it all the way.

## 7 comments:

solution:

let the number be like this:

xn..x2x1x0y

where xn .. x2, x1, x0 are the number digits

Sum (xi * 10^i) + 9999 = Sum (xi * 10^(i+1)) + y, where i = 0..n

this gives:

9 * Sum(xi * 10^i) + y = 9999

now, y can be 0,1,2, .. 9

which means that

Sum(xi * 10^i) = 9999 / 9, 9998 / 9, 9997 / 9, ... 9990 / 9

but Sum must be integer since xi is integer also, hence

999z / 9, z = 0..9 must be integral

so we need to check all 9 ratios

actually its simpler(aratare!) if u divide

9 * Sum(xi * 10^i) + y = 9999

by 9

and u get

Sum(xi * 10^i) + y/9 = 1111

y can be 0 or 9 ;)

Problem #2:

The bridge

Three people (A, B, and C) need to cross a bridge. A can cross the bridge in 10 minutes, B can cross in 5 minutes, and C can cross in 2 minutes. There is also a bicycle available and any person can cross the bridge in 1 minute with the bicycle. What is the shortest time that all men can get across the bridge? Each man travels at their own constant rate.

I didn't solve this problem.

Hint: if you know that d = v * t then everything is easy :|

Solution to problem #2

----------------------

The bicycle can be left at any place on the bridge and can be used in both directions on the bridge.

The steps are:

B and C walk by foot and A take the bicycle.

A at one place (y) can leave the bicycle and finish the way by foot.

C can take the bicycle, drives it back to a point (x), leaves it and continues to walk by foot.

B takes the bicycle and drives it all the way.

The equations are:

A: 1*y + 10*(1-y)

B: 5*x + 1*(1-x)

C: 2*y + (y-x) + 2*(1-x)

The system becomes:

10 - 9y = -3x + 3y + 2 = 4x + 1

Solve each equality:

10 - 9y = -3x + 3y + 2 -> 3x - 12y = -8

10 - 9y = 4x + 1 -> 4x + 9y = 9

After solving the equations:

x = 12/25, y=59/75.

It meas that it takes 73/25 = 2.92 minutes to cross the bridge.

A speed = 1/10

B speed = 1/5

C speed = 1/2

(bicycle speed is 1)

Problem #3:

You have 2 buckets with water, 7 and 4 litre.

How can you measure 1 litre using these buckets ?

Solution to problem #3

----------------------

Fill the 4 litre bucket then fill the 7 litre bucket with it completely.

Now you have 3 litre left in the 7 litre bucket.

Again, fill the 4 litre bucket and fill the remaining 3 litre from the 7 litre bucket.

What you have left in the 4 litre bucket is 1 litre.

problem #4:

exchange 2 variables without using any intermediate value.

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