nakulendu
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Posts posted by nakulendu
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Seven.
Tetrahedron - 4 faces (triangles)
The Square Pyramid - 5 faces (4 Triangles + 1 square)
Glue two together One triangle face to another. so two faces out fo total 9 hence seven - 6 triangles plus 1 square -
Unless i'm missing something, the boy is 3.5 years old
OP says ... "The struggle went on until the class was filled up...."
Class could have filled up on Day 4, on Day 5 (in which case the 3.5 yrs answer is good. But if the class filled up any time later then the answer is not corect.
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If the last lady joined to join did so on Day 4 then the age of 1st girl is 1.75 yrs and 1st Boy 3.5. Actually works out like this:
Last girl to join was on Day 2 - Age of 1st girl - 7 , age of 1st Boy 14 yrs
Last girl to join was on Day 4 - Age of 1st girl -1.75, age of 1st Boy 3.5 yrs
Last girl to join was on Day 6 - Age of 1st girl -.4375, age of 1st Boy .875 yrs
Last girl to join was on Day 8 - Age of 1st girl - .109375, age of 1st boy .21875 yrs
and so on.....
If you imagine that the poor school master did not run a post natal child care unit and given that on day 4 an elder sister was brought in, most logical answer looks like 1.75 as age of 1st girl and age of 1st boy 3.5 and the class filled up on day 4 itself or the next day. -
10 pairs in all.
6 of the pairs are truly 2 digit number pairs:
1. (12, 42 ) (21, 24)
2. (13, 62 ) (31, 26)
3. (14, 82 ) (41, 28)
4. (23, 64) (32, 46)
5. (24, 84) (42, 48) ... already listed
6. (34, 86) (43, 68)
4 additional the pairs can be had where there is a 1 digit number in each pair e.g. you take 1 as 01 and the reversing digits you get 10
). These are
a. (1, 20) (10, 2)
b. (2,40) (20,4)
c. (3,60) (30,6)
d. (4,80) (40,8)
I guess this is it.
Didn't get into 3 digit numbers because rules of switching digits as shown in example gets cloudy
pure 2 digit pairs - 12
and tohose which have a 1 digit number as well as above - 4
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10 pairs in all.
6 of the pairs are truly 2 digit number pairs:
1. (12, 42 ) (21, 24)
2. (13, 62 ) (31, 26)
3. (14, 82 ) (41, 28)
4. (23, 64) (32, 46)
5. (24, 84) (42, 48) ... already listed
6. (34, 86) (43, 68)
4 additional the pairs can be had where there is a 1 digit number in each pair e.g. you take 1 as 01 and the reversing digits you get 10
). These are
a. (1, 20) (10, 2)
b. (2,40) (20,4)
c. (3,60) (30,6)
d. (4,80) (40,8)
I guess this is it.Didn't get into 3 digit numbers because rules of switching digits as shown in example gets cloudy
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Group 1: 18,9,3,6
Group 2: 27, 4,5
both add up to 36 -
Year 2024.
Has Seven instances when product of date and month is the year (modulo 100)
1.24.24
2.12.24
3.08.24
4.06.24
6.04.24
8.03.24
12.02.24 -
if white moves first
1. White : Nxe6 , Black: Nxe6 (otherwise White's next move is the end)
2. White: N a6 , Black can do whatever it wishes
3. White N b8-end -
"I" is... er.... AM the most different
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Box labeled "Gold coins Or Silver Coins" contains Bronze.
Label "Gold" contains Silver and vice versa.
Only this way all labels are incorrect. -
eight if perfectly pointed or absolutely new. Can be no limits when in use (under microscopic examination if required). That is assuming no chips/ dents on the body otherwise.
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OK, five spoons numbered 54,56,58,59,62. Take two at a time.
How ? I got it a bit heuristically. Reasoning being.
1. number of Spoons + number picked each time is ODD. so one of these two numbers is even and the other odd.
2. There are in all 10 different combinations. If all sums were to be unique , you could achieve the 10 combinations by 5 C 2.
3. So the first attempt with 5 numbers - two at a time.
4. spread between the max and min sum is 11, so if this would imply generally a difference of 2 between numbers
5. Among the sums there are 4 odds and 6 even. having one odd number among 5 would make it convenient to achieve this
6. So make one aberration on the Arithmetic series by inseritng a difference of 1 somewhere, immediately followed by a jump of 3 in the series.Generally above to start with, and a little bit of trial and error - QED.
The only point of OP, if at all, one may argue : "She then erases any duplicates...". The above solution has no duplicates. But may be the OP may be interpreted as " She then erases duplicates if any ..."
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the do function "=" is sum up the digits and deduct. So devil's ticket if you please. - 666
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One of them running on a Conveyor belt moving at 10 m/s
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I picked p in my draw. Lets estimate n = p. What would have been the likelihood that I got the max number ? : 1/p
If say n = p+1. What would have been the likelihood I picked p in my draw ? : 1/(p+1)
1/p > 1/(p+1).
So given that I picked p , the estimate of n that would give the best probability of me picking p is p itself.
So best estimate of n is p-
2
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Outward run on trolley : max they can ride for time = 1 hr 54 min 32.73 sec,
Distance - 15.273 mi
then return walk for duration : 5 hr 5 min 27.27 sec would give Mr Smalleash the biggest bang for his buck - in terms of ride time on trolley and time with his girl.However if the girl was not interesting he could always get down earlier and walk back while still being in time to get her home. He could then even save a few pennies on trolly ride as well
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The Only non contradictory case is :
1 is true - in that case 7 = Susan and 8 = Bill
Made an error both 7 and 8 are lying (only 1 is speaking truth. So 7 = Bill and 8 = Susan
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The Only non contradictory case is :
1 is true - in that case 7 = Susan and 8 = Bill -
For Every die I can choose there exists one which has a 66.67 : 33.33 chance to win. e.g. if Dice are named A,B,C,D in order then respectively for each the nemeses with the above mentioned odds are :
For A, it is D ; for B it is A ; for C it is B ; and for D it is C. So not much fair play in that. OF course I will take you on IF :
A. you choose your die first and I choose mine next.
B. I choose first and I prevent you from choosing one of the remaining 3. In that case I will choose C and stop you from taking B. -
Henry does the maths correct. He figures that if the first spin landed an empty as the first shot, the probability that next one would not be empty is 50% (as Tammie reasoned). And on a fair spin the chance of getting an empty is 66.67%. So spin he should go for.
But he knows Gretchen had been itching to get back at him. All the Russian roulette stuff was designed to extend his mental torture. And in fact he was surprised that she had missed on the first spin to land a chamber with bullet under the hammer. Would he give her another chance with a spin or take his 50 odd.
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Bmad runs 100 m in the time Bonanova runs 70.
BMAD's Speed / Bonanova Speed = 100 / 70
Similarly , Bonanova's Speed / Phase's Speed = 100 /80
So BMAD's speed / Phase's speed = 10000/5600 = 100/56
So when BMAD reached finishing line at 100m, phase was at 56 m and hence 44 m behind -
the distance is determinate with given facts if you consider a different kind of a belt. Say the belt is a girdle or metal hoop which does not stretch to form a tangent with your shrunk circumference but stays a larger circle touching your current circumference behind. tried to attach a picture in the file
what you are looking for is the difference in the radii Old - new . (R-r).
Which is 5/2 PI.
or 0.795774715 inches -
Imagine Circular table of radius r. Then cousins will be at vertices of a 9 sided regular polygon inscribed in the circle. The respective distances in question become .Me to Amber (salt travels) = D1 = 2r COS (10o){if you draw the polygon an put people around it in order then Amber and I will be at the base vertices of an Isoscles triangle which has r as the sides and 10o being the base angles. The distance Amber-Me = D1 is the base of this triangle}Amber to Chris (Salt travels) = D2 = 2r COS (50o) {similarly Amber and Chris are also a the base of another Isoscles triangle with equal sides r and the base angle being 50o }Amber to Becky (pepper moves) = D3 = [r COS (50o)]/COS (20o) {try the geometry - Amber - Becky also at base of another isoscles triangle with base angle of 70o. This time we need to take the cosine of the difference 70o- 50o to get the distance D3.}D1 = 1.969615506024416118733486049179 rD2 = 1.2855752193730786526452868198145 rD3 =0.68404028665133746608819922936452 rD1 = D2 + D3
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Mortgage 1296; Electricity 72; Phone Bill 38; Cable Bill 34; Trash collector's bill 18; Paper Boy 6.
Building a better rectangle
in New Logic/Math Puzzles
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