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Ek gali me bachche khel rahay the

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Dimag chalao aur jawab do.
Ek Gali Mein _____ Bacche Khel Rahay The. Uss Gali Mein Se ek _____ Aurat Guzar rahi thi. Uss _____ Aurat ke Hath Mein _____ Thi. Un _____ Bachon Ne Uss _____ Aurat Ki _____ Gira Di. In 7 Khali Jaga Main Ek Hi Word Aye Ga.. Dimag chalao aur jawab do.

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Number Puzzle #68, 1, 10, 30, ?, 130

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Number Puzzle #68

missing number in 1, 10, 30, ??, 130

missing number in 1, 10, 30, ??, 130

Check your answer:-

Tried enough already?


Click here to See Solution

Actually this is a GP with first element as 1 and variable factor. where the factor comes from this series
1, 3, 7, 13.

Tn = Tn-1 + 2^(n-1);
T1 = 1;
T2 = T1 + 2^(2-1)= 1 + 2 = 3;
T3 = T2 + 2^(3-1) = 3 + 4 = 7;
T4 = T3 + 2^(4-1) = 7 + 6 = 13

nth number in GP = first number * (10)(Tn-1 in series);
First Element in GP = 1;
Second Element in GP(i.e. n = 2) = 1*10*(T1 in series); = 10
Third element in GP(i.e. n =3) = 1*10*(T2 in series) = 1*10*3 = 30
Fourth element in GP (i.e. n =4) = 1*10*(T3 in series) = 1*10*7 = 70
Fifth element in GP (i.e. n =5) = 1*10*(T4 in series) = 1*10*13 = 130

Thus the missing number here is 70.

Number puzzle #67, 4+5+4+5=81

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Number Puzzle #67

If 4+5+4+5=81 5+6+5+6=16 6+7+6+7=64 1+2+1+?=16 What is the missing number?

If
4+5+4+5=81
5+6+5+6=16
6+7+6+7=64
1+2+1+?=16
What is the missing number?

Click here to See Answer
First lets understand the pattern
4+5+4+5 => 18 => 1+8 => 9 => square of 9 => 81
5+6+5+6 => 22 => 2+2 => 4 => square of 4 => 16
6+7+6+7 => 26 => 2+6 => 8 => square of 8 => 64

Thus the patter is sum of the digits till we get single digit and then square of the the single digit.

Method 1:
So lets assume the missing number is x

thus 1+2+1+x = x + 4
Now if x is less then or equal to 5(i.e. x+4 is single digit)
(x+4)(x+4) = 16 => x = 0

Now if x is greater then 5( i.e. x + 4 is two digit number)
lets replace x by 6+y (where y is single digit)

x+4 => 6+y+4 => 10 + y => y +1 => (y+1)*(y+1) = 16 => y = 3

thus x = 6+y = 9.

Note: If we assume missing number can be more then single digit.

Method 2:

its the reverse engineering method => to get 16 the sum of digits should be 4

we already have 1+2+1 = 4 thus missing number can be 0
other ways to get the sum of digits as 4 => 13, i.e. missing number = 13-4 = 9
other ways is 22, i.e. missing number 22-4 = 18
other way is 31, i.e. missing number 31 – 4 = 27
other way is 40, thus missing number 40 -4 = 36

so there can be more such numbers

Seven thieves and diamonds

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There are seven thieves, They steal diamonds from a diamond merchant and run away in jungle. While running, night sets in and they decide to rest in the jungle When everybody’s sleeping, two of the best friends get up and decide to distribute the diamonds among themselves and run away. So they start distributing but find that one diamond was extra. So they decide to wake up 3rd one and divide the diamonds again …..only to their surprise they still find one diamond extra. So they decide to wake up fourth one. Again one diamond is spare. 5th woken up……still one extra. 6th still one extra. Now they wake up 7th and diamonds are distributed equally.

How many minimum diamonds they steal?

Check your answer:-

Tried enough already?
See Solution
Lets say there are x diamonds, now these diamonds are exactly divisible by 7.
and
x = 1 + N1*2;
x = 1 + N2*3;
x = 1 + N3*4;
x = 1 + N4*5;
x = 1 + N5*6;
x = N6*7;
where N1, N2, N3, N4 and N5 are integers.
From above we can also say
N1*2 = N2*3 = N3*4= N4*5 = N5*6 = y

Now y should be divisible by 2, 3, 4, 5 and 6, its nothing but common multiple of all.

LCM of 2, 3, 2*2, 5, 2*3 => 2*3*2*5 => 60

at the same time common multiple + 1 should be divisible by 7 as well.

60 + 1 is not divisible by 7
120(60*2) + 1 is not divisible by 7
180(60*3) + 1 is not divisible by 7
240(60*4) + 1 is not divisible by 7
300(60*5) + 1 is divisible by 7

Thus they must have stolen minimum 301 diamonds.

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