@hatemonger said:Approach pls
x + 1/x = 4
squarring both sides,
x^2 + 1/x^2 = 14
subtracting 2 from both sides..
x^2 + 1/x^2 - 2 = 12
(x-1/x)^2 = 12
(x-1/x) = 2_/3
now x^4 - 1/x^4 = (x-1/x) (x+1/x) (x^2 + 1/x^2)
=> 2_/3 * 4 *14= 112_/3
@ScareCrow28 said:P = (2-1)1! + (3-1)2!+ (4-1)3! + .....P = 13! - 1P+3 = 13! + 2Hence remainder by 13! = 2 ?
P = (2-1)1! + (3-1)2!+ (4-1)3! + ..... to P+3 = 13! + 2 , could u pls explain
70^5340 has its right most non zero digit as??? pls explain how to get non zero digit
@Calvin4ever said:P = (2-1)1! + (3-1)2!+ (4-1)3! + ..... to P+3 = 13! + 2 , could u pls explain
Kya explain karu isme ?
@hatemonger said:70^5340 has its right most non zero digit as??? pls explain how to get non zero digit
70^5340
Non zero digit ( last ) = 7^5340 mod 10 = 1
@hatemonger said:70^5340 has its right most non zero digit as??? pls explain how to get non zero digit
f(70^5340) = f( 7^5340)
7 has cyclicity 7,9,3,1
5340 = 4k
so 1
@bodhi_vriksha said:The answer should be 106 .. You guys are missing a small thing ... Identify it And dont forget to post your figures as well Team BV-- Pratik Gauri
yup kamal sir did a little mistake in translating the man from south to east
final equation will be (4h^2-24) +300^2 = (5h-30)^2
h comes out as 106
@hatemonger said:70^5340 has its right most non zero digit as??? pls explain how to get non zero digit
simply set aside 10^5340 and check for last digit in 7^5340
5340 mod 4 =0
last digit is 1
@Dexian said:wo circle wale ka kafi ganda sa answer aa raha hai... wat is OA...
26 cm2. it comes out neat. Draw a rectangle such that OE is one of the diagonal. and apply pythaogorean couple of times.
@Buck.up said:Is Height of Light house 100 m ? If yes then I will attach the solution
Yaar answer is 106 m
@albiesriram said:26 cm2. it comes out neat. Draw a rectangle such that OE is one of the diagonal. and apply pythaogorean couple of times.
i think i m doing that only ......... i ll check again...
@albiesriram said:What is the largest two digit prime factor of 200 C 100?
it will be 61
200!/ 100! 100!
now u need to find the highest prime number that contributes a power of 3 in numerator will a power of 2 in the denominator
for 61 it is 61^3/61^2 =61
for next prime 67 it is 67^2/67^2
67*3=201 so 200/67 =2
67*3=201 so 200/67 =2
@Dexian said:i think i m doing that only ......... i ll check again...
y2+(6-x)2 = x2 + (y+2)2 = 50
where OE = x and OD = y EBDO is the rectangle
@abhishek.2011 said:bhai 67*3 = 201 hota h so i guess 67 will have a power of 2 in 200!
Haan bhai 😞 Ans will be 61.. I need to revise my multiplication process
@abhishek.2011 said:yup kamal sir did a little mistake in translating the man from south to eastfinal equation will be (4h^2-24) +300^2 = (5h-30)^2h comes out as 106
Dude can u attach an answer fig. I am not able to visualise