Arun Sharma Test: Number System- 2 - CAT MCQ

Given (10²⁵−7)−(10²⁴+x) is divisible by 3
10²⁵−7−10²⁴−x
10²⁴(10−1)−7−x
9.19²⁴−7−x
If divisible by 3 value of x=2

Therefore, the correct option is B

=x                                     
► All the digits written once = 1762125.125=10000x
► All the digits without Bar written once = 1762.125=10x
Rational form = 

► There are 4 case in the same
► Case 1 : a and b both are odd

⇒ a² + 2ab + b²
⇒ Odd² + 2 x Odd x Odd +Odd²
⇒ Odd + Even + Odd
⇒ Even

►Case 2 : a and b both are even

⇒ a² + 2ab + b²
⇒ Even² + 2 × Even × Even + Even²
⇒ Even + Even + Even
⇒ Even

► Case 3 : a is odd and b is even

⇒ a² + 2ab + b²
⇒ Odd² + 2 x Odd x Even + Even²
⇒ Odd + Even + Even
⇒ Odd

► Case 4 : a is even and b is odd

⇒ a² + 2ab + b²
⇒ Even² + 2 × Even × Odd + Odd²
⇒ Even + Even + Odd
⇒ Odd

► In 2 Cases its Odd and in other 2 it is Even, so we cannot say anything about the answer. 

► There are two possibilities for a
► Case 1: a is even

⇒ 6a  = Even x Even = Even
⇒ 6a + 12 = Even + Even = Even

► Case 2: a is odd

⇒ 6a = Even x Odd = Even
⇒ 6a + 12 = Even + Even = Even

► In both the cases the answer is even, so the information provided is inconsistent.

Therefore, the last two digits of the multiplication 122×123×125×127×129 are: 50

► For abc to be Odd the only case possible is none of them being Even i.e. a, b and c all three are Odd

⇒ a² + b² + c²
⇒ Odd² + Odd² + Odd²
⇒ Odd + Odd + Odd
⇒ Odd

Vice versa can also be true

► There are 12 prime numbers between 25 and 75 as listed below: 

► There is only one Even prime number i.e. 2. 

► We know the property that a prime number when divided by 6 gives a remainder of 1 or 5.
► Option A : 8831 divided by 6 gives a remainder of 5
► Option B : 8829 divided by 6 gives a remainder of 3
► Option C : 8833 divided by 6 gives a remainder of 1, but is divisible by 11
► Option D : 8835 divided by 6 gives a remainder of 3

► Option A : a² + b² = Odd² + Odd² = Odd + Odd = Even
► Option B  : a² + 2b² = Odd² + Even × Odd² = Odd + Even = Odd
► Option C ∶ 2a² + b² = Even × Odd² + Odd² = Even + odd = Odd
► Option D ∶ a² + b² + ab = Odd² + Odd² + Odd × Odd = Odd + Odd + Odd = Odd

► Sum of Any Consecutive 50 Whole Numbers

⇒ Sum of 25 Even Numbers + Sum of 25 Odd Numbers
⇒ Even + Odd
⇒ Odd

► First 10 Prime numbers are listed Below :

► Their sum = 129

► Being Prime numbers any number would be divisible by 1 or itself.
► Since 17 is a prime number. It will be the only prime number less than 1000 which will be divisible by 17.

► Being Prime numbers any number would be divisible by 1 or itself.
► Since 16 is not a prime number. There would be no other prime number divisible by 16. 

► The List of such pairs is listed below:
(3,5); (5,7); (11,13); (17,19); (29,31); (41,43); (59,61); (71,73), (101,103); (107,109); (137,139);  (149,151); (179,181); (191,193); (197,199) 

► ​15 Such pairs exist.