The only even numbers in the list are 2 and 4, but 4 is not a prime. So 2 can be used to illustrate the statement that all primes are not odd.
When the larger cube is cut into smaller cubes, the corner cubes will have paint on three sides. The cubes in the middle of the faces will have paint on only one side, but the cubes cut from the edges will have paint on two sides. In this case, there will be only one cube one each edge (excluding the corners), and since there are 12 edges, there will be 12 cubes with paint on two sides.
Given a x a = 12
asked to find a x a x a x a = (a x a) x (a x a)
a4 = a2 x a2
= 12 x 12
= 144.
In the given options,
12/49 can be written as 1/(49/12) = 1/4.083
7/30 can be written as 1/(30/7) = 1/4.285
13/56 can be written as 1/(56/13) = 1/4.307
11/46 can be written as 1/(46/11) = 1/4.1818
Here in the above,
1/4.083 has the smallest denominator and so 12/49 is the largest number or fraction.
Any number which can be expressed as a fraction of two integers like P & Q as P/Q where Q is not equal to zero.
Every integer is a rational number since Q can be 1.
Hence, in the given options, 4 can be expressed as a simple fraction as 4/1. And all other options cannot be expressed as fractions.
Hence, 4 is a rational number in the given options.
Let assume the value of a,b &c that will satisfy the equation.
so a=1,b=1,c= -1 or a=1,b= -1 &c=1 or a= -1,b=1,c=1
here all three cases satisfy the equation
assume any one a=1,b=1,c= -1
Hence (a+b)(b+c)(c+a)=(1+1)(1-1)(-1+1) = 0
we can treat every two consecutive terms as one.
So, we will have a total of 100 terms of the nature:
(2 + 5) + (7 + 6) + (12 + 7).... => 7, 13, 19,....
We know the sum of n terms
Now, a= 7, d=6 and n=100
Hence the sum of the given series is
S= 100/2 x[2 x 7 + 99 x 6]
=> 50[608]
=> 30,400.
On dividing 627349 by 15, we get remainder = 4.
Therefore, the obtained remainder is the least number to be subtracted from the given number so that the the mnumber is divisible by 15.
Here 4 is the least number to be subtracted from 627349 so that it is divisible by 15.
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.
(3x + 2) (2x - 5) = ...............(1)
But (3x + 2)(2x - 5) = ........(2)
so by comparing (1) & (2),
we get a= 6, k= -11 , n= -10
(a - n + k) = 6 + 10 - 11 = 5.
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
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