Last for today from my side: Nimai and Nitai start running simultaneously from opposite ends of a road towards each other with speeds of 3m/s and 5m/s respectively. As soon as they reach some end, they turn without stopping and keep running towards other end till both of them reach their start position simultaneously. Nitai hands over a lotus flower to Nimai everytime while meeting him from the opposite direction.And whenever Nitai overtakes Nimai i.e. crossing while moving in same direction, Nitai takes a lotus flower back from Nimai. What is the number of lotus flowers with Nimai at the end of the run?Team BV - Kamal Lohia
since the length of the track is immaterial here so we assume length of the track to be a number which is multiple of 3,5,(5-3) and (5+3) = 120 m faster guy hits the start point every 240/5 = 48 secs, slower guy hits the end every 240/3 = 80 secs
Meet head to head: now every time the sum of distances covered by them is an odd multiple of 120 they will meet each other while approaching hence they will meet each other approaching after every 120(2k+1)/(5+3) = 15, 45, 75, ..... secs
Overtake: every time the difference of distance covered is an odd multiple of 120 faster guy overtakes the slower guy hence every 120*(2k+1)/(5-3) = 60,180,300 secs the flower is taken back.
Reaching the start point at same time(race ends): This happens when both the guys at the same end
hence LCM of (48,80) = 240 sec
=> flower will be given after every 15 secs. (i.e 15, 45, 75, 105, 135, 165,195,225.... secs) i.e. 8 times
flower will be taken back at 60 and 180 secs
hence 8 -2 = 6 flowers.
Apologies for incorrect solution earlier, i had misread the question and made an error.
How many sets of three distinct factors of the number N = 2^6 — 3^4 — 5^2 can be made such that the factors in each set has a highest common factor of 1 with respect to every other factor in that set?
How many sets of three distinct factors of the number N = 2^6 — 3^4 — 5^2 can be made such that product of these factors is equal to N and HCF of these factors(in particular set) is equal to 1 ?
A 12 centimeter cube (i.e. 12x12x12) was painted red on the outside. The cube was then cut into one centimeter cubes. How many of these smaller cubes have paint on one side? Two sides? Three sides? No sides?
VibhorThere is nothing bold in your above post.Anyway, it says for a number N, how many sets of three co-prime factors (a, b, c) can be formed?For example for 6 we have only one such set (1, 2, 3) but for 120, we have (1, 2, 3), (3, 4, 5) and many more.I hope it is clear now. Team BV - Kamal Lohia
Ohhh... I thought we have to make sets such that the product of factors give the number itself! Thank you for the reply 😃 I made the Un-Bold part Bold after editing..
How many sets of three distinct factors of the number N = 2^6 — 3^4 — 5^2 can be made such that product of these factors is equal to N and HCF of these factors(in particular set) is equal to 1 ?
all must be co-prime, and prodcut should be N
now, one set is 2^6, 3^4, 5^2
others can be N as a product of 2 co-prime numbers and 1 = 2^2 = 4
How many sets of three distinct factors of the number N = 2^6 — 3^4 — 5^2 can be made such that product of these factors is equal to N and HCF of these factors(in particular set) is equal to 1 ?
A 12 centimeter cube (i.e. 12x12x12) was painted red on the outside. The cube was then cut into one centimeter cubes. How many of these smaller cubes have paint on one side? Two sides? Three sides? No sides? pls give the details of solution
One Side: All surface cubes - edges cubes = 10*10*6 = 600
Two Sides: Edges cubes = 10*12 = 120
Three sides: Corner cubes = 8
No side = Total - (1+2+3 sides painted) OR (as done by the birdie below) = 12^3 - (728) = 1000 ? ..
A 12 centimeter cube (i.e. 12x12x12) was painted red on the outside. The cube was then cut into one centimeter cubes. How many of these smaller cubes have paint on one side? Two sides? Three sides? No sides? pls give the details of solution
one side only= 10*6 *10=600 two sides only=10*12=120 [cubes on the edges-corner cubes] 3 sides only=8 [corner cubes] no sides = 10*10*10 = 1000 [scrape off one layer and you get clean cubes!] basically,for no cubes (n-2)^3 :)
A 12 centimeter cube (i.e. 12x12x12) was painted red on the outside. The cube was then cut into one centimeter cubes. How many of these smaller cubes have paint on one side? Two sides? Three sides? No sides?
pls give the details of solution
Let's take the general case of a cuboid, with dimensions a*b*c.
a*b*c can be expressed as ((a-2) +2)*((b-2) +2)*((c-2) +2) which on simplification gives:
(a-2)(b-2)(c-2)+ 2{(a - 2)(b - 2)+(a - 2)(c - 2)+(b - 2)(c - 2)} + 4(a - 2) + 4(b - 2) + 4(c - 2) + 8 which will give u answer for sides painted nil,one, two and three respectively.
Whenever in doubt just deduce this expression, and again for cube it is real easy, just make a=b=c=L
((L-2)+2)^3= (L-2)^3 + 3*(L-2)^2 *2 + 3*(L-2)*2^2 + 2^3 which will give u answer for sides painted nil, one, two and three respectively.
For, L=12 the values are 1000, 600, 120, 8 respectively
Let's take the general case of a cuboid, with dimensions a*b*c.a*b*c can be expressed as ((a-2) +2)*((b-2) +2)*((c-2) +2) which on simplification gives:(a-2)(b-2)(c-2)+ 2{(a - 2)(b - 2)+(a - 2)(c - 2)+(b - 2)(c - 2)} + 4(a - 2) + 4(b - 2) + 4(c - 2) + 8 which will give u answer for sides painted nil,one, two and three respectively.Whenever in doubt just deduce this expression, and again for cube it is real easy, just make a=b=c=L((L-2)+2)^3= (L-2)^3 + 3*(L-2)^2 *2 + 3*(L-2)*2^2 + 2^3 which will give u answer for sides painted nil, one, two and three respectively.For, L=12 the values are 1000, 600, 120, 8 respectively Team BV -Vineet
Thanks buddy...ur technique is awsome..really helpful
A 12 centimeter cube (i.e. 12x12x12) was painted red on the outside. The cube was then cut into one centimeter cubes. How many of these smaller cubes have paint on one side? Two sides? Three sides? No sides? pls give the details of solution
A 12 centimeter cube (i.e. 12x12x12) was painted red on the outside. The cube was then cut into one centimeter cubes. How many of these smaller cubes have paint on one side? Two sides? Three sides? No sides? pls give the details of solution
I thought we have to make sets such that the product of factors give the number itself! Thank you for the reply 😃 I made the Un-Bold part Bold after editing..
