@ScareCrow28 said:Find all nonnegative integers n such that there are integers a and b with the propertyn^2 = a + b and n^3 = a^2 + b^2.
a=0,b=0,n=0
a=1,b=0,n=1
a=2,b=2,n=2
could only find 3
@ScareCrow28
@ScareCrow28 said:Find all nonnegative integers n such that there are integers a and b with the propertyn^2 = a + b and n^3 = a^2 + b^2.
n^2=a+b
n^3=a^2+b^2
==> n^4 = a^2 +b^2 +2ab
n^4=n^3 +2ab
n^4-n^3=2ab
n^3(n-1) = 2ab
(a^2+b^2)(n-1)=2ab
(a^2+b^2)/(ab)=2/(n-1)
==> (a/b + b/a) = 2/(n-1)
now RHS and LHS will be equal for only 1 value when a=b = 2 for which n= 2
for any value of n greater then 2 RHS will be
PS: using this method a=0, b=0, n=0 is not feasible solution
@anytomdickandhary said:There would be many such sets of 10 integers.Take any 10 multiples of 3 (say a1,a2,a3.......a10)Say the sum of the squares of these numbers is Snow you can generate 10 distinct integers using formulaXi = S/9 - ai^2These ten integers generated will satisfy the required conditions.One such set of integers is.376.00 349.00 304.00 241.00 160.00 61.00 -56.00 -191.00 -344.00 -515.00 ATDH.
Very well explained .. But then isn't it easy to do like
first 10 integers sum 55 .. ignore 6 .. 7^2
first 12 integers sum 78 ignore 14 ( 4,7,3 or 6,7,1) .. 8^2
so many sets
first 10 integers sum 55 .. ignore 6 .. 7^2
first 12 integers sum 78 ignore 14 ( 4,7,3 or 6,7,1) .. 8^2
so many sets
@naga25french said:Very well explained .. But then isn't it easy to do likefirst 10 integers sum 55 .. ignore 6 .. 7^2first 12 integers sum 78 ignore 14 ( 4,7,3 or 6,7,1) .. 8^2so many sets
I think we have comprehended the problem a bit differently. The way I comprehended this is that if we have a set of 10 distinct integers, then if we pick any 9 of these integers and sum it then it should be a perfect square.
for example as suggested by you first 10 integers sum is 55. So if we select first 9 then we get sum as 45 which is not a perfect square. Where as in the set that I suggested sum of any nine integers would give you a perfect square.
Please correct me if I am wrong any where.
ATDH.
@anytomdickandhary said:Please correct me if I am wrong any where. ATDH.
Nah not wrong .. Solution is perfect 
@bullseyes said:@ScareCrow28n^2=a+bn^3=a^2+b^2==> n^4 = a^2 +b^2 +2abn^4=n^3 +2abn^4-n^3=2abn^3(n-1) = 2ab(a^2+b^2)(n-1)=2ab(a^2+b^2)/(ab)=2/(n-1)==> (a/b + b/a) = 2/(n-1)now RHS and LHS will be equal for only 1 value when a=b = 2 for which n= 2for any value of n greater then 2 RHS will be PS: using this method a=0, b=0, n=0 is not feasible solution
I have problems with your solution. I have highlighted that in your original post.
You have divided the RHS by (n-1). You can't do that unless it's given that n is not =1
So, here by doing this you have ignored 1 value of n, i.e, n=1
Secondly, You have divided the LHS by 2ab, so again you have negated the values a=b=0 by doing this.
I suppose you have neglected two cases of n. Your previous answer was correct.
A faulty clock gains 10 minutes every hour. If the time is set correctly at 12 Noon on 1st Jan 2010,
then how many times will its minute-hand and hour-hand meet in the next 24 hours ?
(a) 22 (b) 26 (c) 24 (d) 25
then how many times will its minute-hand and hour-hand meet in the next 24 hours ?
(a) 22 (b) 26 (c) 24 (d) 25
@sparklingaubade said:A faulty clock gains 10 minutes every hour. If the time is set correctly at 12 Noon on 1st Jan 2010,then how many times will its minute-hand and hour-hand meet in the next 24 hours ?(a) 22 (b) 26 (c) 24 (d) 25
Relative speed of faulty minute hand=60+10-5=65 points per hour
Total points covered for one meet =60 points
So for one meeting time taken=60/65
So it 24hrs number of meetings=24*65/60=26 times...Is it OA?
Total points covered for one meet =60 points
So for one meeting time taken=60/65
So it 24hrs number of meetings=24*65/60=26 times...Is it OA?
@TootaHuaDil
Nopes
There are three watches 창€“ W1, W2 and W3. Once an alarm goes off in W1, it rings continuously for
20 seconds, then pauses, then starts ringing again after 2 minutes, and so on. The respective
values for W2 are 50 seconds and 5 minutes, and for W3 are 1 minute and 6 minutes. An alarm is
set in each of the three watches for 06:00 am. At what time after 06:00 am will the three alarms go
off simultaneously for the first time?
(a) 06:35 am (b) 06:42 am (c) 06:30 am (d) None of these
20 seconds, then pauses, then starts ringing again after 2 minutes, and so on. The respective
values for W2 are 50 seconds and 5 minutes, and for W3 are 1 minute and 6 minutes. An alarm is
set in each of the three watches for 06:00 am. At what time after 06:00 am will the three alarms go
off simultaneously for the first time?
(a) 06:35 am (b) 06:42 am (c) 06:30 am (d) None of these
@sparklingaubade said:A faulty clock gains 10 minutes every hour. If the time is set correctly at 12 Noon on 1st Jan 2010,then how many times will its minute-hand and hour-hand meet in the next 24 hours ?(a) 22 (b) 26 (c) 24 (d) 25
d) 25?
60 minutes of normal clock = 70 minutes of faulty clock
24* 60 minutes of normal clock = 24*70 minutes of faulty clock
i.e. 240 extra minutes or 4 extra hours
now minute hand and hour hand coincide 11 times in 12 hours
so in 28 hours they coincide 28*11/12 = 77/3 = 25.66
So 25 times
60 minutes of normal clock = 70 minutes of faulty clock
24* 60 minutes of normal clock = 24*70 minutes of faulty clock
i.e. 240 extra minutes or 4 extra hours
now minute hand and hour hand coincide 11 times in 12 hours
so in 28 hours they coincide 28*11/12 = 77/3 = 25.66
So 25 times
@sparklingaubade said:A faulty clock gains 10 minutes every hour. If the time is set correctly at 12 Noon on 1st Jan 2010,then how many times will its minute-hand and hour-hand meet in the next 24 hours ?(a) 22 (b) 26 (c) 24 (d) 25
The faulty clock gains 10 minutes every hour, so weed to calculate the no of times hands meet in 24*70 minutes.
In 24*60 minutes they meet 22 times, hence in 22*70 minutes they would meet 25 times...
.
@ScareCrow28 said:Find all nonnegative integers n such that there are integers a and b with the propertyn^2 = a + b and n^3 = a^2 + b^2.
(a+b)^2=n^4
n^3=(a+b)^2-2ab
n^4-n^3=2ab
n^3(n-1)=2ab
n={0,.2,4,6,8,10--------------all even numbers}
n={1,3,5,7,9----------------all odd numbers }
@gnehagarg said:(a+b)^2=n^4n^3=(a+b)^2-2abn^4-n^3=2abn^3(n-1)=2abn={0,.2,4,6,8,10--------------all even numbers}n={1,3,5,7,9----------------all odd numbers }
Just check those values and see whether a and b are coming out to be integers and satisfying
n^2 = a + b and n^3 = a^2 + b^2 or not.
Ans is 3 values
@sparklingaubade said:@TootaHuaDilNopesThere are three watches €“ W1, W2 and W3. Once an alarm goes off in W1, it rings continuously for20 seconds, then pauses, then starts ringing again after 2 minutes, and so on. The respectivevalues for W2 are 50 seconds and 5 minutes, and for W3 are 1 minute and 6 minutes. An alarm isset in each of the three watches for 06:00 am. At what time after 06:00 am will the three alarms gooff simultaneously for the first time?(a) 06:35 am (b) 06:42 am (c) 06:30 am (d) None of these
(a) 06:35 am?
Ton + Toff
W1 : 1/3 + 2 = 7/3 = 14/6
W2 : 5/6 + 5 = 35/6
W3 : 1 + 6 = 7 = 42/6
LCM(14/6,35/6,42/6) = 210/6 = 35 minutes
So 6.35 am
Ton + Toff
W1 : 1/3 + 2 = 7/3 = 14/6
W2 : 5/6 + 5 = 35/6
W3 : 1 + 6 = 7 = 42/6
LCM(14/6,35/6,42/6) = 210/6 = 35 minutes
So 6.35 am
@sparklingaubade said:@TootaHuaDilNopesThere are three watches €“ W1, W2 and W3. Once an alarm goes off in W1, it rings continuously for20 seconds, then pauses, then starts ringing again after 2 minutes, and so on. The respectivevalues for W2 are 50 seconds and 5 minutes, and for W3 are 1 minute and 6 minutes. An alarm isset in each of the three watches for 06:00 am. At what time after 06:00 am will the three alarms gooff simultaneously for the first time?(a) 06:35 am (b) 06:42 am (c) 06:30 am (d) None of these
W1 --- 20s + 120s = 140sec
W2----50sec + 300 sec = 350 sec
W3 ---60 + 160 = 420 sec
LCM (140, 350, 420) = 2100
So, after 2100 sec or 35 Minutes they ring again simultaneously
Time 6:35 a.m.
@ScareCrow28 said:I have problems with your solution. I have highlighted that in your original post.You have divided the RHS by (n-1). You can't do that unless it's given that n is not =1 So, here by doing this you have ignored 1 value of n, i.e, n=1 Secondly, You have divided the LHS by 2ab, so again you have negated the values a=b=0 by doing this.I suppose you have neglected two cases of n. Your previous answer was correct.
ahh.. tht will eventually lead to 3 solutions.
@ScareCrow28 said:The faulty clock gains 10 minutes every hour, so weed to calculate the no of times hands meet in 24*70 minutes.In 24*60 minutes they meet 22 times, hence in 22*70 minutes they would meet 25 times....
Please let me know my conceptual flaw. I know I am making mistake. But where? 
@TootaHuaDil said:Relative speed of faulty minute hand=60+10-5=65 points per hourTotal points covered for one meet =60 pointsSo for one meeting time taken=60/65So it 24hrs number of meetings=24*65/60=26 times...Is it OA?
@sparklingaubade said:@TootaHuaDilNopesThere are three watches €“ W1, W2 and W3. Once an alarm goes off in W1, it rings continuously for20 seconds, then pauses, then starts ringing again after 2 minutes, and so on. The respectivevalues for W2 are 50 seconds and 5 minutes, and for W3 are 1 minute and 6 minutes. An alarm isset in each of the three watches for 06:00 am. At what time after 06:00 am will the three alarms gooff simultaneously for the first time?(a) 06:35 am (b) 06:42 am (c) 06:30 am (d) None of these
6:35?
LCM (140,350,420) = 2100 seconds = 35 minutes
@TootaHuaDil said:Please let me know my conceptual flaw. I know I am making mistake. But where?
According to your solution, the hands of the clock meet 24 times in a day. But, that is not the case. The hands , in fact, meet 22 times in a day. So , I think that is the flaw in your solution.
@TootaHuaDil said:Please let me know my conceptual flaw. I know I am making mistake. But where?
the minute hand and the hour hand don't meet every 60 minutes.. they meet every 65 5/11 minutes.. :)
So it will be (60*24)/(65 5/11 * 60/70 * 24) = 25.667 :D
So it will be (60*24)/(65 5/11 * 60/70 * 24) = 25.667 :D