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How many handshakes in African-US summit ?

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During the African-US summit last year, 10 presidents left the conference room and greeted one another. Assuming no 2 presidents repeatedly greeted themselves, how many total handshakes were effected?

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Answer is 45 handshakes. The first president greeted 9 presidents, the second 8 and so on, so 9+8+7+6+5+4+3+2+1. This gives 45 handshakes.

in permutations and combinations way
A handshake is possible with two people, so it is similar to How many ways you can select two people out of 10.

i.e. 10C2 = 10!/(8!*2!) = 9*10/2 = 45

Farmer, 3 sons and chapatis

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A farmer’s wife made some chapatis…the farmer had 3 sons…. The fist son came, gave one chapati to the dog,and made three equal parts of remaining chapatis, ate one part of it and left the other two parts for his brothers….other two sons came one after the other and did the same thinking that they came first…then at night all three came to the house, one of them gave one chapati to dog and made three equal parts and the three brothers ate one-one part of it… If no chapati was broken in pieces then how many minimum number chapatis did the mom made?

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Lets say their mom made x chapatis
number of chapatis left after first son: 2(x-1)/3
number of chapatis left after second son: 2(2(x-1)/3 – 1)/3 = (4(x-1) – 6)/9 = (4x -10)/9
number of chapatis left after third son: 2((4x -10)/9 -1)/3 = (8x – 20 – 18)/27 = (8x – 38)/27
they ate ((8x-38)/27 -1)/3 chapatis each in the night

So it should be an integer value, lets say it is y, where y is an integer.
(8x-38 – 27)/81 = (8x – 65)/81 = y
x = (81*y+ 65)/8 = 10y + (65+y)/8 = 10y + 8 + (y+1)/8
as x should also be an integer
y can be => 7, 15, 23, ….
x can be = 79, 160, 241

minimum value of x is 79.

Find the value of (A+B)

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A*(4+(B/100))=108.81
find the value of (A+B) by solving the above equation

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A*(4+(B/100)) = 108.81

4A + AB / 100 = 10881/100

4A + (AB/100) = 108 + (81/100)

So,

4A = 108 and AB/100 = 81/100 => AB = 81

Hence

A=108/4=27 and by solving AB 81, B = 3
Thus A+B = 30

This solution is courtesy Shashikant Soni

Four digit number in aabb

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There is four digit number in aabb form and it is a perfect square. Find out the number.

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We can expand aabb as

=> 1000a + 100a + 10b + b
= 1100a + 11b = 11(100a + b)

for 11(100a +b) to be a perfect square 100a + b should be a multiple of 11 where a and b both are integers and single digit numbers.

lets say 100a + b = 11y, here y should also be a perfect square (such that 11*11*y is a perfect square), i.e. 1, 4, 9, 16, 25, 36

If you try to solve this, where a and b are single digit integers and y is also a perfect square, you can see one possible values of a and b are 7 and 4, where y = 64.

Thus the number should be 7744, it is a perfect square of 88.

Four rivers and three temples

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Puzzle Time:
Four rivers and three temples

All the rivers are magical, moment a worshiper crosses a river with any number of flowers, it becomes double. (For eg 1 flower becomes 2 flowers, 2 become 4 and so on).

How many minimum number of flowers a worshiper needs 2 carry from beginning such that he offers equal number of flowers in all the three temples, and he is left with zero flowers at the end of fourth river.

Solve this ??

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Lets assume that he offers x flowers at every temple, thus he must have only x flowers left at the end of 3rd temple, otherwise he will not have zero flowers at the end of fourth river.

also we can say that after offering flowers at second temple he should be left with x/2 flowers.
similarly after offering flowers at first temple he should be left with (3x/2)/2 flowers.
Thus he should be carrying (x + 3x/4)/2 flowers initially, i.e. 7x/8 flowers initially.

For number of flowers to be integer, x =8, thus he must carry 7 flowers initially and offer 8 flowers at each temple.

Ultimate math puzzle

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Solve this

0 0 0=6
1 1 1 = 6
2 2 2=6
3 3 3 =6
4 4 4=6
5 5 5=6
6 6 6=6
7 7 7=6
8 8 8=6
9 9 9=6

In the space you can use any kind of functions like addition subtraction divison etc
But you cannot bring up new numbers
Eg: 2+2+2=6

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0) (0! + 0! + 0!)!
1) (1+1+1)!
2) 2+2+2=6
3) 3×3-3=6
4) 4+4-( square root of 4)=6
5) (5÷5)+5=6
6) 6+6-6=6
7) 7-(7÷7)=6
8) (cube root of 8) + (cube root of 8) + (cube root of 8) = 6
9) (square root of 9)* (square root of 9)- (square root of 9)=6

Solve 36 % 6 x 3 -9 + 1

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Kids math puzzle #1 36%6*3-9+1 = ?

Kids math puzzle #1
36%6*3-9+1 = ?

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Here as there are no braces, operators will work as per their precedence.
division and multiplication has same precedence and addition and subtraction has same precedence. Also / and * has higher precedence then + and -.
Thus first % and * should operate, now they being of the same precedence, they will work from left to right order thus % will operate first and the * will work on its result.

Thus 36%6*3 -9 +1 => 6*3 -9 +1 => 18 -9 + 1=> 10

Arrange 7 men in alternative height

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There are 7 people numbered as 1 to 7.
The number denotes their height.
Arrange them in a line in such a manner so that the line appears in alternative of their heights, I.e. one short then long, again short then long.
The condition is that, if one by one member(remove 1, then 2, then 3) from the line is omitted, the order of their height remains alternative.

When do you have Rs.60

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You are depositing Rs. 3 in First day and Rs. 3 in Second day in your account. And you are withdrawing Rs. 4 from your account in Third day. You do this repeatedly. When soon do you have Rs. 60?

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After third day you will have Rs.2 in your account (3+3-4). So your account will gain Rs.2 every 3rd day.

Thus you would have reached Rs. 60 in (60/2)*3 = 90 days, but important thing to notice is that you are withdrawing Rs. 4 on every third day so your balance will reach Rs.60 mark earlier than that.

So, if we observe we will reach (60-6)=Rs.54 in (54/2)*3=81 days, and as you will deposit Rs.6 in next two days, you will reach Rs.60 in 83days.

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