@Tusharrr
sorry for wrong qstn
correct question is
what is the largest factor of 11! that s one bigger than a multiple of 6 ?
@nits_rocker said:@Tusharrrsorry for wrong qstncorrect question iswhat is the largest factor of 11! that s one bigger than a multiple of 6 ?
385?
7*11*5
@nits_rocker said:@AsihekAdhvaryuanswer s correct....but plz explain in detail
Answer should be in the form of 6k+1 so it must be odd
possible values 3 5 7 9 11
3^3,5^1,7^1,11^1
check remainder for 6
3,5,1,5 or 3,-1,1,-1 (to make remainder 1 just multiply 1*-1*-1 =1)
but We can not multiple 3 3^2 or 3^3 as it will make the remainder 3
Though not a very good way but 
A group of 5 people are going to meet weekly at the library for 4 weeks. Each week, two people are selected at random to speak. Each person may speak in multiple weeks, but no pair of people will speak together more than once. The probability that there is a person who will never be asked to speak can be expressed as a/b where a and b are coprime positive integers. What is the value of a+b ?
Plzzzzzzz solve
if 20 engineer and 20 worker can together construct 20 km road in 20 days. 40 engineer and 40 worker together construct 40 km road in how many days?
@Tusharrr 7 got 10 pairs
So no of ways =10 × 9 × 8 × 7
No of ways after 1 person excluding group = 6 group
I did nt got next
@vbhvgupta
Let speed of eng. Be e and worker be w
20 days * [ (20 e )Km/day + (20 w) Km/day ] = 20 km
20e+20w = 1
E+w=1/20
Next
[40 (e+w)] * x days = 40
(E+W) * X = 1
1/20 * X =1
X =20 DAYS
@vbhvgupta said:if 20 engineer and 20 worker can together construct 20 km road in 20 days. 40 engineer and 40 worker together construct 40 km road in how many days?
20
@Tusharrr said:A group of 5 people are going to meet weekly at the library for 4 weeks. Each week, two people are selected at random to speak. Each person may speak in multiple weeks, but no pair of people will speak together more than once. The probability that there is a person who will never be asked to speak can be expressed as a/b where a and b are coprime positive integers. What is the value of a+b ?Plzzzzzzz solve
There are total C(5, 2) = 10 pairs
Out of these 10 pairs, 4 can be chosen in C(10, 4) ways.
Since one of the person has never been called we will be left with C(4, 2) = 6 pairs
Out of these 6 pairs, 4 can be chosen in C(6, 4) ways
=> Required probability = C(6, 4)/C(10, 4) = 15/210 = 1/14
a + b = 15
@nits_rocker said:@Tusharrrsorry for wrong qstncorrect question iswhat is the largest factor of 11! that s one bigger than a multiple of 6 ?
Factor of the form 6k+1
Remainder by 6 should be 1
11! has 5 odd numbers namely; 3, 5, 7, 9 and 11 ( Excluding 1 of course )
Remainders = 3, -1, +1, +3, -1
Possible Combination = (-1)(1)(-1)
Number = 5*7*11 = 385 .. ?
@nits_rocker said:@Tusharrrsorry for wrong qstncorrect question iswhat is the largest factor of 11! that s one bigger than a multiple of 6 ?
11! = (2^8)(3^4)(5^2)(7)(11)
Factor is of form 6k + 1, so it will not have 2 or 3 as a factor
Now, to get 6k + 1, we need to multiply factors of form 6p + 1 (any number of times) and factors of form 6p - 1 (even number of times)
=> out of 5, 5, 11, we can pick atmost two and since we have to find largest such factor we should choose 5 and 11
=> Largest such factor = 5*7*11 = 385
@ScareCrow28
since we need to find largest factor why cant we do power of any of 5*7*11
y cant the answer be 5^2*7*11?
since we need to find largest factor why cant we do power of any of 5*7*11
y cant the answer be 5^2*7*11?
@nits_rocker said:@ScareCrow28since we need to find largest factor why cant we do power of any of 5*7*11y cant the answer be 5^2*7*11?
That way the factor won't be of the form 6k+1, instead now it is of the form 6k-1
@vbhvgupta said:if 20 engineer and 20 worker can together construct 20 km road in 20 days. 40 engineer and 40 worker together construct 40 km road in how many days?
20 days
