# Question Bank - Number Theory - Shashank Prabhu, CAT 100 Percentiler

• The answer is not B. You can draw a circle of any dimension and join 2 points on the circumference that are separated by 8. So, the max possible is infinite.

• We can all possible groups of 1 per person each OR all possible groups of 2 persons each OR all possible groups of 3 persons each OR all possible groups of 4 persons each. These will be 4c1, 4c2, 4c3 and 4c4 respectively. If the weights of the four friends are a, b, c and d respectively, and if we consider one person at a time, the total weight of all possible combinations will be a+b+c+d. For combinations of groups of 2 people each, it will be (a+b)+(a+c)+(a+d)+(b+c)+(b+d)+(c+d). For all groups of 3 people it will be (a+b+c)+(a+b+d)+(a+c+d)+(b+c+d). For four people it will be a+b+c+d. As the weight of each person is less than 100, exactly one person groups and exactly four persons groups are out as the total will be less than 400. For all possible groups with exactly 2 people, we get 3a+3b+3c+3d=882 which is possible and so, average as 73.5. For all possible groups with exactly 3 people we again get 3a+3b+3c+3d=882 and the average as 73.5.

• Q3) Tarun invited 750 people to a party. Everyone who came to the party, came by a car. Three cars carried 6 people each and rest of the cars carried 7 people each. Only those people came to the party who were invited by Tarun. At the party, when every person including Tarun was seated in groups of eleven, one group fell short by four people. If maximum possible number of people came to the party, then out of the people who were invited by Tarun, how many people did not come to the party?

[OA: 18]

• Should be 7n + 4 = 11m + 6, 39 is the remainder when divided by 77.
So 77k + 39. 732 is the highest value less than 750.
So 18 didn't attend.

• Q5) Which of the following cannot be expressed as the sum of three distinct composite numbers?
(1) 21
(2) 23
(3) 16
(4) 18

[OA: Option 3]

• Q9) There are 49 zeros, 51 ones and 53 twos written on the board randomly. A student is blindfolded and then asked by his teacher to touch any two numbers on the board arbitrarily. The teacher deleted those two numbers and replaced them by a single number in the following manner:

The pair is replaced by

(0, 0) → 0
(1, 1) → 0
(2, 2) → 2
(1, 2) → 1
(0, 1) → 1
(0, 2) → 0

If they continued this process what was the number left on the board in the end?

(a) 0
(b) 1
(c) 2
(d) Cannot be determined

[OA: Option B]

• After the first iteration: Wine will be 2/4 = 3/6
After the second iteration: Wine will be 1.5/5 = 3/10
After the third iteration: Wine will be 1.2/6 = 3/15
After the fourth iteration: Wine will be 1/7 = 3/21
Can you see the pattern?

• Q17) How many four-digit numbers with distinct digits are there such that the sum of the digits is even?

[OA: 2376]

• Q18) Given that a^5 + b^5 + c^5 = 91849, where a, b, and c are distinct digits. What is the remainder when a six-digit number 2a5b1c is divided by 11?

[OA: 0]

• 55x + 68 = 56(x+1), x=12.
Total was 660 and 12 elements.
Sum of first 11 will be 66 at minimum.
So remaining is 594.

• Q21) There are three natural numbers X, Y, and Z such that the LCM of (X, 120) is 1320, LCM of (Y, 120) is 1680 and LCM of (Z, 120) is 1800. Which of the following statements is true?

a. X, Y, and Z all three can be perfect squares.
b. Only Y and Z can be perfect squares.
c. Z can be a perfect square.
d. Z is definitely a perfect square.

[OA: Option c]

• Need to understand that the lines are parallel to each other.
Then simply distance between the two parallel lines and distance between P and Q should give you area of the triangle.

• 2 cases: first 3 digits consecutive; for 123, 234, 345, 456, 567, 678 last digit could be chosen in 9 ways.
For 789 it could be chosen in 10 ways.
last 3 digits consecutive; for 012, 123 in 9 ways, for 234, 345, 456, 567, 678, 789 8 ways each.
Total of 64+18+48=130

• Q29) What is the sum of all those factors of 1800 whose units digit is 5 ?

[OA: 390]

• If there were x from Dakistan and y from Pagladesh, we would have xc3=455 and yc3=120. x=15, y=10. So, 25 people in total. Now, we have to take the total possible number of teams and subtract teams with only Dakistanis and Pagladeshis in them. So, 25c3-15c3-10c3=1725

• Q37) If the last digits of the products 1 * 2, 2 * 3, 3 * 4, ..., n * ( n+1) are added, the result is 2010. How many products will be used to get this result?
a. 1000
b. 1005
c. 1010
d. Cannot be determined

[OA: Option d]

• Q41) The number of persons who booked a ticket for the New Year's concert is a perfect square. If 100 more persons had booked a ticket then the number of spectators would be a perfect square plus 1. If still 100 more persons had booked a ticket then the number of spectators would be again a perfect square. How many persons booked ticket for the concert?

[OA: 2401]

• Let the length be x. The first part if increased to 1.5 times will become x/3. So, original length of the first part will be 2x/9. If the second part becomes 2/3rd of itself, it will be x/3. So, original length of the second part will be x/2. The third part will be x - 2x/9 - x/2 = 5x/18.

• Q44) Find the sum of all remainders when n^5 – 5n^3 + 4n is divided by 120 for all positive integers n > 2017.

[OA: 0]

• Let Ramesh be 40 m/s, Shyam be 10 m/s. Ramesh reaches 1200 m after 30 seconds. Shyam covers 300 m in that time. New speed of Ramesh is 5 m/s, distance between Ramesh and Shyam is 700 m. So, they meet after 140 seconds. Shyam covers 1400 m more in this time period. So, total of 1900 m from the start or 100 m from the end

47

138

31

145

200

68

61