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

 

Let
x = peacocks
y = deers

2x + 4y = 200
x + y = 72
y = 72 - x
2x + 288 - 4x = 200
x = 44 peacocks
y = 28 deers

Number = quotient x divisor + remainder;
so, here
number = 138 k + 26
=> (23 x 6k) + (23+3)
=> 23(6k+1)+3
so, remainder is 3.

For solving this problem first we would break the whole range in 5 sections

1) From 1 to 9
Total number of zero in this range = 0

2) From 10 to 99
Total possibilities = 9*1 = 9 ( here 9 is used for the possibilities of a non zero integer)

3) From 100 to 999 - three type of numbers are there in this range
a) x00 b) x0x c) xx0 (here x represents a non zero number)
Total possibilities
for x00 = 9*1*1 = 9, hence total zeros = 9*2 = 18
for x0x = 9*1*9 = 81, hence total zeros = 81
similarly for xx0 = 81
total zeros in three digit numbers = 18 + 81 +81 = 180

4) From 1000 to 9999 - seven type of numbers are there in this range
a)x000 b)xx00 c)x0x0 d)x00x e)xxx0 f)xx0x g)x0xx
Total possibilities
for x000 = 9*1*1*1 = 9, hence total zeros = 9*3 = 27
for xx00 = 9*9*1*1 = 81, hence total zeros = 81*2 = 162
for x0x0 = 9*1*9*1 = 81, hence total zeros = 81*2 = 162
for x00x = 9*1*1*9 = 81, hence total zeros = 81*2 = 162
for xxx0 = 9*9*9*1 = 729, hence total zeros = 729*1 = 729
for xx0x = 9*9*1*9 = 729, hence total zeros = 729*1 = 729
for x0xx = 9*1*9*9 = 729, hence total zeros = 729*1 = 729
total zeros in four digit numbers = 27 + 3*162 + 3*729 = 2700
thus total zeros will be 0+9+180+2700+4 (last 4 is for 4 zeros of 10000)
= 2893

(3x + 2) (2x - 5) = ax2+kx+n ...............(1)

But (3x + 2)(2x - 5) = 6x2-11x-10........(2)

so by comparing (1) & (2),

we get a= 6, k= -11 , n= -10

(a - n + k) = 6 + 10 - 11 = 5.

By trial and error method, we get
2880/3 = 960 is not a perfect square
2880/4 = 720 is not a perfect square
2880/5 = 576 which is perfect square of 24
Hence, 5 is the least number by which 2880 must be divided in order to make it into a perfect square.

Intial contribution = L/K
After 3 men drop, then contribution =L/K-3
The amount more to pay in contribution = LK-3 - LK

= 3LK(K-3)

Let the smaller number be k.
Then, larger number = (3355 + k)
Therefore 3355 + k = (6k + 15)
‹=› 5k = 3340
‹=› k = 668.

Therefore, the smaller number k = 668 and now the larger number = 3355+668 = 4023.

Given number is 987 = 3 x 7 x 47.

So, required number must be divisible by each one of 3, 7, 47.

None of the numbers in 553681 and 555181 are divisible by 3. While 556581 is not divisible by 7.
Correct answer is 555681.