# Quant Boosters - Hemant Malhotra - Set 2

• let base is n
61 * 61 = 4541
(1 + 6n) (1 + 6n) = 1 + 4n + 5n^2 + 4n^3
1 + 12n + 36n^2 = 4n^3 + 5n^2 + 4n + 1
so n = 8

• Q28) What is the remainder when 27^11 + 9^11 + 3^11 + 1 is divided by 3^10 - 1

• 3^10 = x
so 27 * 3^30 + 3^2 * 3^20 + 3 * 3^11 + 1
so 27 * x^3 + 9x^2 + 3x + 1 put x = 1
so 27 + 9 + 3 + 1 = 40

• Q29) What is the last digit of the LCM of 7^5678 - 1 and 7^5678 + 1 ?

• HCF = 2
LCM * HCF = first number * second number
LCM * 2 = ( 7^5678 - 1) * (7^5678 + 1)
unit digit of 7^5678 = 9
so unit digit of 7^5678 - 1 will be 8
unit digit of 7^5678 + 1 will be 0
so unit digit of LCM will be 0

2nd method to solve this
LCM * 2 = (7^5678 - 1) * (7^5678 + 1)
= 49^5678 - 1
(49-1)(1 + 49 + 49^2 + ... 49^5677)/2 = 24 (1 + 49 + 49^2 + ... 49^5677)
49^k mod 10 = 1 when k is even and is -1 when k is odd
now check inside portion is zero

• Q30) Seven Hockey teams A, B, C, D, E, F and G participated in a tournament. Each team played with every other team twice and D won all of its games and F lost all of its games. B and G scored equal points and are ahead of C. C, A and E also scored equal number of points. Each team gets 3 points, 1 point and no point for a win, draw and loss respectively. What is the highest possible number of points that can be scored by team A?

• Total number of matches = 7C2 * 2 = 42
If there were no matches that resulted in a draw then total points = 42 * 3 = 126
For every match resulting in a draw the total points get reduced by 1.
Because in a Win-Loss situation Winner + Loser = 3 + 0 = 3 points but in a draw it is 1 + 1 = 2 points.
Let the total points of A,C,E be y and that of B and G be x
3y + 2x + 36 = Total points [ also y < x ]
Total points can be 126 or 125 or 124 and so on
But our objective is to maximize y we have to take value of total as high as possible.
3y + 2x = 126 then max = 16
but if we take 3y + 2x = 125 then max(y) = 17 so OA=17

• @hemant_malhotra Kindly explain the solution, I couldn't really comprehend it.

• @Aayushiv This is a way of solving sets without venn diagrams.
Please go through the below notes which explains the method in great detail. Once you are done, re-visit the problem and if the doubt persist, let us know. Will help you (I guarantee, it will be clear once you are done with the 2 notes given below!) :)

• @hemant_malhotra shouldn't A point at the top so the distance become 2+3+2 = 7
as in room A at top face not at bottom face

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