How many ways can you put 10 sweets into 3 bags so that each bag contains an odd number of sweets?
1@Adithya, there are more for 4.....
15 112 113
8 15 17
how to do this ?? what logic did you use ???
269for the bag question..i was just thinking that we can do it lyk this....take the 1st bag and put 2 sweets in it now put the 2nd bag in it with 5 sweets in it...so..now the first bag contains=2+5=7 sweets....
next, we are left with only 3 sweets...so put them in the third bag...n keep it separate....in the way,,we have all three bags..with odd no. of sweets...
Astha Gupta u have already answered it..but can we do lyk this..???..
114.the logic is-
we have to find integral solution for the eq.s
a2+b2=152
a2-b2=152
separately.
32+42=52 so 92+122=152
so in the first case the only soln is (9,12,15)
in the 2nd case
a2-b2=(a+b)(a-b)=225.1=75.3=45.5=25.9
solve for a+b=225,a-b=1 (113,112,15)
a+b=75,a-b=3 (39,36,15)
a+b=45,a-b=5 (25,20,15)
a+b=25,a-b=9 (17,8,15)
SATISFIED?:-)
1057but dont u think then the question becomes too easy...has to be some other interpretation.... :)
1057but @aditya it isnt mentioned dat de next palindromic no is displayed.....
2626. the next palindomic number is 16061
distance = 16061-15951 = 110km
time = 2hrs
avg speed = 110/2 = 55km/h
1057@johncenaiit if wat u said is allowed...
den surely in all the combinations 2 bags will have odd no of sweets and 1 bag will have even no of sweets....
therefore we have to find integral solutions of x+y+z=10 where x and y are odd and z is even...
replacing x, y and z with 2k+1,2l+1 and 2m we get...
k+l+m=4
where k ,l,m can be 0 also....
i say m can be zero because z is the bag which has a bag inside and it may not necessary have a sweet of its own...
so no of non-negative integral solutions of k+l+m=4 is 6C2 which is 15... :)
2625. i) one side = no. of face centers of cube =6
ii) two sides = no. of small cubes sharing side with face center= 12
iii) three sides = no. of corners =8
iv) no side = central cube = 1
2621. after every twenty counts,the process repeats itself.
now 1982 = 20k +2
hence counting ends on the 2nd finger.
16.A car milometer shows 15951. After 2 hours it shows another palindromic number. How fast was the car going on average?
1try this too............
1.Holding its hands out, palms upward a child starts counting on all its fingers and thumbs, going to and fro. If it counts up to 1982 which finger does counting end on?
2.You have 3 bricks, each measuring 18 x 9 x 6 cm. How many different heights can you build up with them?
3.If D = the day (1-366) in year Y, then the day of the week can be calculated using
d = D+Y+(Y-1)/4 - (Y-1)/100 + (Y-1)/400 mod7
where d=1 would mean Sunday, etc. Can the first day of each century (e.g. 1st Jan 2001, 1st Jan 1901) be any day?
4.How many right-angled triangles with integral sides have one side of 15?
5.A cube is painted white and cut into 27 small cubes. How many of these little cubes are painted on i) 1, ii) 2, iii) 3, iv) 0 sides?
1i don't know ketan .....:(
but they provide a hint : The first trick is to realise that if you put one bag inside another, then sweets in the inner bag are also in the outer bag. The only workable configuration is to put one bag inside another and leave the third alone
if you got it pls explain to me......
1057arre...@johncenaiit....
if u put odd sweet in each of the three bags how will their sum result in 10...it will be a odd number....
is it necessary to give all the sweets?