Arun Sharma Test: Number System- 3 - Interview Preparation MCQ

= x + y = 12

► First 5, 3 digit prime numbers are listed below:

► Their Sum is 533

► The given terms are: a, a + 2, a + 4
► For Less than 5, We know of such a set: 3, 5 and 7
► For Prime numbers Greater than 5 we have two cases
► Case 1 : The Prime number on division by 6 gives a remainder of 1
• Let us assume a is a prime number
• So a = 6p + 1
• a + 2 = 6p +3. (This will be divisible by 3. Thus not a prime)
• a + 4 = 6p + 7

► Case 2 : The Prime number on division by 6 gives a remainder of 5
• Let us assume a is a prime number
• So a = 6p + 5
• a + 2 = 6p + 7
• a + 4 = 6p + 9 (This will be divisible by 3. Thus not a prime.)

► Hence there is only one such set.

► Prime numbers is a property of natural numbers, There are no negative prime numbers. 

► All the two digit Prime numbers are listed below: 
►The sum is 1043
►The negative of this sum is - 1043

► Option A : b is Even
⇒ 2a + 3b + 9c = 2 x any number + 3 x Even + 9 x any number = Even + Even + Odd/Even
⇒ This does not give a sure shot result.

► Option B : c is Even
⇒ 2a + 3b + 9c = 2 x any number + 3 x any number + 9 x Even = Even + Odd/Even + Even
⇒ This does not give a sure shot result.

► Option D : Either of the two (b or c) is Even.
⇒ 2a + 3b + 9c = 2 x any number + 3 x Even + 9 x any number = Even + Even + Odd/Even
⇒ 2a + 3b + 9c = 2 x any number + 3 x any number + 9 x Even = Even + Odd/Even + Even
⇒ This does not give a sure shot result.

► Option C = Both b and c are Even
⇒ 2a + 3b + 9c = 2 x any number + 3 x Even + 9 x Even = Even + Even + Even = Even
⇒ So only option C gives an even response.

► Highest 3 digit prime number = 997
► Smallest 3 digit prime number = 101
► Difference = 997 - 101 = 896

Solve this question through the options. For n terms being 16 (option 1), we would need an AP with 16 terms and common difference 1, that would add up to 1000. Since, the average value of a term of this AP turns out to be 1000÷16=62.5, we can create a 16 term AP as 55,56,57....62,63,64...70 that adds up to 1000. Hence, a series of16 consecutive terms is possible. Likewise, a series of 5 terms gives the average as 200 & the 5 terms can be taken as 198,199,200,201,202. It is similarly possible for 25 terms with an average of 40 but is not possible for 20 terms with an average of 50.

Hence, option (d) is correct.

► It is +1 with exceptions of prime numbers below 6 obviously.

► Any prime number above 6 is of the form:

6k + 1 or 6k - 1 ( A lot of people don't know this!))

► So therefore squares is of the form (6k ± 1)². Expand this and clearly the only term which is not a multiple of 6 will be 1. Which again makes it of the form 6K + 1( now only + because we squared the whole term, Remember?)!.

► So therefore the remainder is of the form +1.

► So for every prime no. above 6, the remainder 1.

So, for 2, it is 4

For 3, it is 3.

For 5, it is 1.

► For every other prime no. Square, It’s 1.

Let n(a,b) is two digit number.

n(a,b)=10a+b

Given  sum of its digitsnumber​ will give 4 as quotient and 3 as remqinder.

n(a,b)=4(a+b)+3

10a+b=4a+4b+3

6a−3b=3⇒2a=1+b

n(a,b)=3(ab)+5

10a+b=3ab+5

10a+2a−1=3a(2a−1)+5

12a−1=6a2−3a+5

6a2−15a+6=0

2a2−5a+2=0

(2a−1)(a−2)=0

a=2(∵a=21​)

∴b=2a−1=3

23 is required number.

So, the correct option is B.

Divide the given expression by 1000 and find the remainder to get the answer.
(12345 × 54321) ÷ 1000 = (2469 × 54321) ÷ 200 → gives a remainder of (69 × 121) ÷ 200 = 8349 ÷ 200 → gives us a remainder of 149. Thus, the remainder would be 149 × 5 = 745. Hence, the last three digits would be 745

The correct option is C 111..... twenty ones
The required GCD would be 1111...111 twenty ones.

The remainder of (1010+10100+101000+....+1010000000000)÷7→(310+3100+31000+.....+310000000000)÷7→(34+34+34+34+34+34+34+34+34+34)÷7=Remainder of 40÷7→5.

So, the correct option is C

All prime numbers between 1 and 20 = 2, 3, 5, 7, 11, 13, 17, 19

Perfect square numbers = 4, 9, 16 and perfect cube numbers = 8 should not be picked.

Remaining numbers respectively. 6, 10, 12, 14, 15, 18, 20. Out of these, 12 and 18 gives perfect squares when multiplied with 3 and 2

1 has to be kept in the list since perfect squares and cubes are left out (taking two perfect squares would result in their product being a perfect square too). Perfect cube has been dropped off the list since its multiplication with 2 would give a perfect square.

Hence, list of numbers = 1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 20

Thus, there are total 14 such numbers.

Hence, 14 is the correct answer.

Properties of 9

The remainder we get when we divide the sum of the digits by 9 is the same as the remainder when we divide the number by 9.

1+6+1+7 = 15

remainder = 1+5=6

Hence, the required answer is 6.

To maximize number of unique question you have to simultaneously decrease the duplication of the question. Hit and trial method. 1st 12 catalyst has 50 unique question = 12 × 50

= 600

But we can have only at the maximum 35 unique question per catalyst. Hence, next 12 catalyst question from 1st 12 catalyst = 35 x 12 = 420

As we can only have 35 unique question. Hence, the last catalyst will have the 35 unique question.

600+420 +35 1055

Hence, the required answer is 1055.

Greatest number of 4 digits = 9999

For any number to be divisible by 345, it should be divisible by 5, 3, 23.

9999 is divisible by 3, but not divisible by 5 and 23

When we divide 9999 by 23, it leaves a remainder 17

So, if we add [i.e., 6 17 + 6 = 23] to 9999

The result will be exactly divisible by 5 and 23, [i.e., 9999 + 6 = 10005]

So, the least number to be added is 6 and the number obtained is 10005.