The greatest possible number which can always divide the sum of the cubes of any three consecutive integers is _____
If (pqr)^2 = (ijkpqr) where i, j, k, p, q,r E of W. , and pqr and ijkpqr are three digits and 6 digits numbers respectively. Then the value of i*j*k*p*q*r is :
@Calvin4ever said:The number of co-primes of 200 lying between 1 to 100 is ____
Is it 40??
200 = 2*2^2*5^2 = 2^3*5^2
100*1/2*4/5 = 40...
@ishu1991 said:bhai its solution is far too lonf see i m posting itSuppose the result is not true. Then if all guests simultaneously move either 0,1,2,…15 places to the right (say), at most one person will be in their proper place. Because each of the sixteen guests come to their proper place at least once in this procedure, it must be the case that after each of these 16 moves, exactly one person will be in their proper seat. If we then determine, for each person the “distance” from their proper seat (all measured from the beginning position, in the direction to their right), we get each of the numbers 0,1,2,…,15 one time. Imagine now, that one of the guests goes (to their right) to the proper seat. Then imagine that the person in that place goes (always to the right) to the proper place., and continuing, each person going to their proper place after getting “bumped” by another, until eventually one of the persons in the chain arrives at the empty chair vacated by the first guest. Each person in this chain moves their own “distance” (defined above), and the total of the distances in this chain of moves will be a multiple of 16 ( because the chain ends at the same place in which it started).If this “chain reaction” procedure moves each guest to their proper place we can continue the arrangement. Otherwise, repeat the procedure another time, starting with one of the guests who is not properly seated. Once again, the sum of the “distances” moved in this sequence of moves is a multiple of 16. Repeat this procedure until everyone is in the proper place. We already have observed that the sum of the “distances” for all the guests is necessarily equal to 0+1+2+….+15 = 120. On the other hand, we have also seen that it is a multiple of 16. seat (all measured from the beginning position, in the direction to their right), we get each of the numbers 0,1,2,…,15 one time. Imagine now, that one of the guests goes (to their right) to the proper seat. Then imagine that the person in that place goes (always to the right) to the proper place., and continuing, each person going to their proper place after getting “bumped” by another, until eventually one of the persons in the chain arrives at the empty chair vacated by the first guest. Each person in this chain moves their own “distance” (defined above), and the total of the distances in this chain of moves will be a multiple of 16 ( because the chain ends at the same place in which it started).If this “chain reaction” procedure moves each guest to their proper place we can continue the arrangement. Otherwise, repeat the procedure another time, starting with one of the guests who is not properly seated. Once again, the sum of the “distances” moved in this sequence of moves is a multiple of 16. Repeat this procedure until everyone is in the proper place. We already have observed that the sum of the “distances” for all the guests is necessarily equal to 0+1+2+….+15 = 120. On the other hand, we have also seen that it is a multiple of 16. It follows that 120 = for some number . However, this equation has no integer solution. Consequently, our original assumption is not possible, so such a simultaneous move is possible. The proposition need not hold if the number of people is 15 instead of 16. An example of this is shown below where the first set of numbers shows the place card numbers, the second the numbers “attached” to each guest and the third line denote each person's distance from their proper place (measured in the number of places they must move to their right). Note that these numbers are around a round table, meaning that number 14 and 0 in the first line are next to each other.placement of guests 8 1 9 2 10 3 11 4 12 5 13 6 14 7 0around the table--------------------------------------------------------------------------------------Place cards 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14--------------------------------------------------------------------------------------each guest's distance 7 0 8 1 9 2 19 3 11 4 12 5 13 6 14from his/her proper placeFor example: guests number 8 is sitting at place card number 0, therefore he should move 7 seats to the right to get to his “correct place card (see the three green numbers in the table above). Guest number 1 is sitting at place card 1, he should move 0 seats to theright to be seated at his place card. Guest number 9 is sitting at place card number 2, he should move 8 seats to the right around the table to seated at his place card, etc. Observe that different guests have different distances, so that regardless of how they rotate as a group, only one person will be properly placed.
how do u have the patience to type ??
@Calvin4ever said:The greatest possible number which can always divide the sum of the cubes of any consecutive integer is _____
is this question complete?? 
@Calvin4ever said:If (pqr)^2 = (ijkpqr) where i, j, k, p, q,r E of W. , and pqr and ijkpqr are three digits and 6 digits numbers respectively. Then the value of i*j*k*p*q*r is :
625^2 ends in 625
625^2 = 390625..
@Logrhythm said:625^2 ends in 625 625^2 = 390625..
How do u directly land in 625 ?? any chain of thought behind this ?
@Logrhythm said:Is it 40?? 200 = 2*2^2*5^2 = 2^3*5^2100*1/2*4/5 = 40...
Is it like co-primes of 200 must be a co-prime of a number below it. Still a bit confused !!
@Calvin4ever said:How do u directly land in 625 ?? any chain of thought behind this ?
1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 6
5^2 = 5
6^6 = 6
7^2 = 9
8^2 = 4
9^2 = 1
10^2 = 0
now we see only 1, 5, 6 and 0 end in the same digit as the number's unit digit...
but 0 and 1 can be easily discarded...
5 and 6 are left...
xx6^2 always has a odd tens digit...
we are left with 5...rest u can figure out easily which square it is...
hope samajh aya hoga..
@Calvin4ever said:Is it like co-primes of 200 must be a co-prime of a number below it. Still a bit confused !!
did not get your point??
@Logrhythm said:1^2 = 12^2 = 43^2 = 94^2 = 65^2 = 56^6 = 67^2 = 98^2 = 49^2 = 110^2 = 0now we see only 1, 5, 6 and 0 end in the same digit as the number's unit digit...but 0 and 1 can be easily discarded...5 and 6 are left...xx6^2 always has a odd tens digit...we are left with 5...rest u can figure out easily which square it is... hope samajh aya hoga..
thanks a lot 😃 crystal now !!@Calvin4ever said:The greatest possible number which can always divide the sum of the cubes of any three consecutive integers is _____
is 36 the number?? did it by observation can be wrong.
@Calvin4ever said:The greatest possible number which can always divide the sum of the cubes of any three consecutive integers is _____
is it 3??
@Logrhythm said:did not get your point??
Is ur solution--> co-primes below 200 - co-primes below 100 ?
@Calvin4ever said:The greatest possible number which can always divide the sum of the cubes of any three consecutive integers is _____
my take 36
@bs0409 said:is it 3??
No, (x-1)^3 + x^3 + (x+1)^3 = 3x (x^2 +2)
Now when x is not divisible by 3, x^2+2 would be divisible by 3. Hence its 9!!
@Calvin4ever said:The greatest possible number which can always divide the sum of the cubes of any three consecutive integers is _____
to find the greatest possible number...take the smallest possible case..
0^3+1^3+2^3 = 9...hence 9..
or this can be done like this..
(a-1)^3+a^3+(a+1)^3 = (a^3+1^3+3a^2+3a)+a^3+(a^3-1^3-3a^2+3a) = 3a^3 + 6a = 3a(a^2+2)
here a>1
hence 3*2 = 9...