Which of the following can be used to illustrate that not all prime numbers are odd? 

 

Let y be the total number of loaves
a be the remaining loaves after 1st thief
b be the remaining loaves after 2nd thief
and c, d..so on after 3 and 4 th
e = 0 be the remaining loaves after 5th thief

a = y - (y/2 + 1/2)
= y/2 - 1/2

b = a - (a/2 + 1/2)
= a/2 - 1/2

c = b - (b/2 + 1/2)
= b/2 - 1/2

d = c - (c/2 + 1/2)
= c/2 - 1/2

e = d - (d/2 + 1/2)
= d/2 - 1/2
we know that, e = 0,
thus by substitution, we get
d = 1
c = 3
b = 7
a = 15
y = 31
Therefore total number of bread loaves in bakery was 31.

Assume that the number is taken as XYZ.
Condition 1 => Z = sq.root(x) => Possible digits of X is 1,4,9
Condition 2 => Y = X + Z
X Y Z
1 2 1
4 6 2
9 12 3 => (wrong because it is 4 digit.)

and 121 is not divisible by 2

The answer is 462.

If he wants to give these books equally to 2 of his friends, then 1 book will remain, it means possible number of books = 15,17,19,21,23,25,27,29
If he wants to give these books equally to 3 of his friends, then 2 books will remain, it means possible number of books = 17,20,23,26,29
If he wants to give these books equally to 4 of his friends, then 3 book will remain, it means possible number of books = 15,19,23,27
common number is only 23 so it will be answer

All powers of 6 has 6 in units place so answer will definitely be 1.

If fares for cities L and M from K are x and y respectively, then

2x + 3y = 77 or 6x + 9y = 231...(1)
3x + 2y = 73 or 6x + 4y = 146...(2)

subtracting 2 from 1, we get

5y = 85

y = Rs. 17 => x = Rs. 13

Let,

no. of rice bowl = x
no. of dal bowl = y
no. of meat bowl = z

Then, x + y + z = 65 ....(1)

Also given that, 2x = 3y = 4z

substituting value of y & z in terms of x i.e. y = 2x/3 and z = x/2 in eq (1),
We get x = 30.

As 1 rice bowl is shared between 2 guests,
Therefore, there are 60 guests.

4, 14, 24, 34, 44, 54, 64, 74, 84, 94 => has 11 number of (4's)
40, 41, 42, 43, 45, 46, 47, 48, 49 => has 9 number of (4's)
So, totally (11 + 9 = 20) number of 4's in 1 to 100.

Sum of decimal places = 7.

Since the last digit to the extreme right will be zero (since 4 x 5 = 20), so there will be 6 significant digits to the right of the decimal point.