If the sum of 101 distinct terms in arithmetic progression is zero, in how many ways can three if these terms be selected such that their sum is zero ?
144?
-50,-49,-48,...,0,...,49,50 0 as the middle term : 50 ways Two positive, one negative: -3,-4,-5,-6,-7,...,-49 => 47 ways Two negative, one positive: 47 ways Total = 50 + 47 + 47 = 144
If the sum of 101 distinct terms in arithmetic progression is zero, in how many ways can three if these terms be selected such that their sum is zero ?
1275??
series -> -50x,-49x,-48x,....0,....48x,49x,50x
case1 - when 0 is picked....we just need to pick 2 complementary tems -> 50 ways
case2 - when 0 is not picked....we need to pick 2 +ve/-ve terms and a terma compensating their weight -> 50c2 ways = 1225
If m is a natural no. and N = 2^m, what is the remainder when N! is divided by 2^N
let m = 2
=> N = 4
(2^2)!%2^4 = 4!%2^4 = 24%16 = 8
if the options are in variables then i can make a few more cases and substitute the values in the options...pl let me know the options if 8 is not the exact answer...
@gnehagarg Please exaplain how did the first staement arrive???b+3a+c+2+1+1+a+3c+b+3+3+2+c+3b+a+1+2+3.. this one..
b+7a+7^2c+2*7^3+7^4*1+3*7^5+7^6*a+7^7*c+7^8*b+7^9*1+3*7^10+2*7^11+7^12*c+7^13*b+7^14*a+7^15*3+7^16*2+7^17*1 Take remainders when divided by 4 you got same result
Consider the series: 1,4,7,10,13,... In the first 300 numbers, how many numbers can be expressed as a difference of two perfect squares in at least one way?
@Logrhythmcase2 - when 0 is not picked....we need to pick 2 +ve/-ve terms and a terma compensating their weight -> 50c2 ways = 1225
I didnt understand case2. Please give some example...
lets pick 2 +ve terms in 50c2 ways
if they are 22 and 24 so their sum is 46 now to counterbalance their weight we need to pick -46 so -46 comes by default or gets picked on its own when we pick the 2 +ve terms....
you can do opposite also by pickng 2 -ve terms and then the +ve terms comes by default...
but i guess there is an error in my answer coz terms like 50+49 (99), 50+48 (98).....we don't have the complementart term for their balancing...
Consider the series: 1,4,7,10,13,... In the first 300 numbers, how many numbers can be expressed as a difference of two perfect squares in at least one way?
Every day asha's husband meets her at railway station at 6:00 pm and drives her to their residence.One day she left early from the office and reached railway station by 5:00 pm. She started walking towards her home,met her husband coming from their residence on the way and they reached home 10 mins earlier than usual time.For how long did she walk.(a) 1 hr (b) 50 mins (c) half hour (d) 55 mins
Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?
let m = 2=> N = 4(2^2)!%2^4 = 4!%2^4 = 24%16 = 8if the options are in variables then i can make a few more cases and substitute the values in the options...pl let me know the options if 8 is not the exact answer...
lets pick 2 +ve terms in 50c2 waysif they are 22 and 24 so their sum is 46 now to counterbalance their weight we need to pick -46 so -46 comes by default or gets picked on its own when we pick the 2 +ve terms....you can do opposite also by pickng 2 -ve terms and then the +ve terms comes by default...but i guess there is an error in my answer coz terms like 50+49 (99), 50+48 (98).....we don't have the complementart term for their balancing...need to remove 25 such cases...answer should be 1275 - 25 = 1250...@bs0409 - pl confirm if the answer is 1250??
OA-1250............
x,y,z are natural numbers whose sum is 12.If maximum value of X*Y^2*z^3 is M, then how many numbers less than M is a co-prime of M?
