@hemant_malhotra Is ans 65?

Q30) Find the number of non negative integral solutions to the equation a + b + c = 13, such that a, b and c are distinct

@hemant_malhotra 256 solutions.

@hemant_malhotra as the no. of real roots is asked so it will always be 3

a^2 - a + 1 mod 13 = 0a = 10^pa(max) = 10^910^18 - 10^9 + 1 mod 13 = 1 + 1 + 1 = 310^16 - 10^8 + 1 mod 13 = 3 - 9 + 1 = 810^14 - 10^7 + 1 mod 13 = 9 + 3 + 1 = 13 mod13=0 sop = 7 N = 16! + 89 mod 323323 = 17 * 19N mod 17 = -1 + 4 = 3N mod 19 = 9 + 13 = 3q = 3 46c23 mod 23 = = 2=r p + q + r = 12

@hemant_malhotra I don't understand how 248 came

999 = 37 * 27 9999 = 99 * 101 999999 = 999* 1001 = 999 * 11 * 91 = 999 * 11 * 13 * 7 2004 is a multiple of 3, 4 and 6. The given number can be divided into groups of three 9's, four 9's or six 9's at a time. The given number is divisible by 7, 13, 37 and 101.

x^2 - 6x + 25 = (x-3)^2 + 16 so min of sqrt (x^2-6x+25) = 4 y^2 - 8y + 25 = (y-4)^2 + 9 so min of sqrt (y^2-8y+25) will be 3 so min sum will be 7 for x=3 and y=4

a + 5b + 10c + 25d = 50 , Put d = 0, a + 5b + 10c = 50 now solve this Put d = 1 then d = 2

Method1- Let's say original weight is 100 now he is taking 20% extra so total 120 and selling 10% less so 90. So (120-90/90) * 100 = 30/90 * 100= 33.33% Method2- multiplying factor1 * 10/9 * 6/5 = 1.33 so 33%

let total work=60 unit 3 M + 5 W per day work=20 unit let M work done in one day=a then work work done=(20-3a)/5 2 M work done in one day=2a and 3 work done in one day=((60-9a)/5 now 60/2a = ((60/((60-9a)/5 -5 so 30/a +5 = 300/(60-9a) find a then ratio

for any weight we have 3 possibilities:- Place it on the weight panPlace it on other pan, along with something whose weight we need to findWe don't use itSo, it is similar to base 3, hence by taking weight of form 3^n, we can ensure that we are using least number of weights and when we just have two possibilities. so 3^n >=300 so n=6

@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

@pratik0809 It is asked to create subset so that NO two elements sum to 11. So Null set satisfies right ? Logic here is, for a usual case (without any conditions), we can have 2^n subsets for a set of n elements. Why ? because each element can either be present or not in the subset. So each element can have two options giving a total of 2^n possibilities. Here, we have 5 pairs and each pair can contribute in 3 ways - First element selected or Second element selected or None is selected. So total subsets = 3^5 = 243 ways (which includes null set too)

@hemant_malhotra please i am not able to solve this do tell me the solution [email protected]

@hemant_malhotra how is it box 2 ???kindly explain

@hemant_malhotra 4!5c25

@hemant_malhotra 40 seconds

@hemant_malhotra 7.4 is the answer !!

@hemant_malhotra 41

@hemant_malhotra said in Question Bank - Algebra - Hemant Malhotra: Let a < b < c be the three real roots of the equation √2014 x^3 - 4029 x^2 + 2 = 0. Find b(a + c). https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_9

@vishwa2017 for question no. 2, 256 is a perfect square, so i think answer should be 14