The number (x y z x y z) can be written, after giving corresponding weightage of the places at which the digits occur, as 100000 x + 10000y + 1000z + 100x + 10y + z
= 100100x + 10010y + 1001z
= 1001 (100x + 10y + z)

Correct Option: D

The number (x y z x y z) can be written, after giving corresponding weightage of the places at which the digits occur, as 100000 x + 10000y + 1000z + 100x + 10y + z
= 100100x + 10010y + 1001z
= 1001 (100x + 10y + z)
Since 1001 is a factor, the number is divisible by 1001.
7 × 11 × 13 = 1001
As the number is divisible by 1001, it will also be divisible by all three namely, 7, 11 and 13
and not by only one of these because all three are factors of 1001.
So, the answer is 1001.

Lets assume the number as N.
Since when this number is divided by 899 it gives us the remainder 63 implies.
N - 63 is divisible by 899.
So the question is what would be the remainder when this number is divided by 29.
Since 899 is multiple of 29.
Implies all the numbers that are divisible by 899 are also divisible by 29.
So N - 63 is also divisible by 29.
But we need what is the remainder when N is divided by 29.
So the answer would be the remainder when 63 is divided by 29.
So the remainder is 5.

Correct Option: B

Lets assume the number as N.
Since when this number is divided by 899 it gives us the remainder 63 implies.
N - 63 is divisible by 899.
So the question is what would be the remainder when this number is divided by 29.
Since 899 is multiple of 29.
Implies all the numbers that are divisible by 899 are also divisible by 29.
So N - 63 is also divisible by 29.
But we need what is the remainder when N is divided by 29.
So the answer would be the remainder when 63 is divided by 29.
So the remainder is 5.
Explanation :
Since N - 63 is divided by 29
Then N - 63 + 29 must be divisible by 29
=> N - 34 is divisible by 29
And then N - 34 + 29 is also divisible by 29
=> N - 5 is divisible by 29
So the remainder will be 5 when N is divided by 29.

Let the two number are a and b
According to the question
sum of squares of two numbers = 97
i.e.., a ² + b ² = 97 .....................(i)
and square of their difference = 25
i.e., (a - b) ² = 25 .................(ii)
⇒ a - b = 5...............(iii)
from Eq. (ii)
a² + b² - 2ab = 25
(a² + b²) - 2ab = 25
Now put the value of a ² + b ² from Eq. (i)
⇒ 97 - 2ab = 25
⇒ 2ab = 72
⇒ ab = 36.................(iv)
Solve the equation.

Correct Option: B

Let the two number are a and b
According to the question
sum of squares of two numbers = 97
i.e.., a ² + b ² = 97 .....................(i)
and square of their difference = 25
i.e., (a - b) ² = 25 .................(ii)
⇒ a - b = 5...............(iii)
from Eq. (ii)
a² + b² - 2ab = 25
(a² + b²) - 2ab = 25
Now put the value of a ² + b ² from Eq. (i)
⇒ 97 - 2ab = 25
⇒ 2ab = 72
⇒ ab = 36.................(iv)
Now, we have
(a + b)² = ( a² + b²) + 2ab
Now put the value of a ² + b ² from Eq. (i) and ab from Eq. (iv)
(a + b)² = 97 + 72 = 169
a + b = 13.............(v)
Now add the Eqs. (iii) and (v), we get
a - b + a + b = 5 +13
2a = 18
⇒ a = 9
now put the value of a in Eq. (v)
a + b = 13
9 + b = 13
b = 13 - 9
b = 4
∴ Product of both the numbers = ab = 9 x 4 = 36

By trial, we find that the smallest number consisting entirely of fives and exactly divisible by 13 is 555555. On dividing 555555 by 13, we get 42735 as quotient.
∴ Req. smallest number =42735.

Correct Option: C

By trial, we find that the smallest number consisting entirely of fives and exactly divisible by 13 is 555555. On dividing 555555 by 13, we get 42735 as quotient.
∴ Req. smallest number =42735.

let the two-digit number be 10p + q
Now according to the question
p + q = 10
and (p +10q) + 36 = (q + 10p)
⇒ -9q + 9p = 36
⇒ -q + p = 4 ....(ii)

Correct Option: B

let the two-digit number be 10p + q
Now according to the question
p + q = 10
and (p +10q) + 36 = (q + 10p)
⇒ -9q + 9p = 36
⇒ -q + p = 4 ....(ii)
on adding eq (i) and (ii) we get
2p = 14 ⇒ p =7
∴ p = 7 and q = 4

∴ Required number is the 73
So, the number is a multiple of prime number.