Hi, I am currently preparing for a maths competitions coming up soon. I received a set of questions to help prepare for it. I can’t solve some of them and it’ll be great you someone could help me out. Working with your solution would be great. Thanks!
1)100 people are standing in a line and they are required to count off in fives as: ‘1,2,3,4,5,1,2,3,4,5,…’ and so on from the 1st person in the line. Anyone who counts ‘5′ must leave the lin. What was the original position in the line of the last person to leave?
2)in the tens digit of a perfect square number is 7, how many units digits are possible. (reasoning please)!
3)When three spherical balls, each of radius 10cm, are placed in a hemispherical dish, it is noticed that the tops of the balls are all exactly level with the top of the dish. What is the radius, in centimetres, of the dish.
4)How MANY ways can you arrange the numbers 1,2,3,4,5,6,7,8,9 in that order with only addition signs. Eg. 1+2+3+45+6+78+9=14
Sorry, for the 4th question, i left out the fact that the numbers must sum up to 144.
clearing up confusion in the 2nd question: after one round of counting to 5, it starts from the beginning again. e.g. once it goes to person 100, it returns to person 1 and the counting begins again.
1)
I had the 47th number left at the end when I used a spreadsheet to actually do this. When I get home this afternoon I will have a think about a way of solving this without actually doing it. If I come up with something I will let you know. Because as this stands I have tried all sorts and I can’t work out how to do it if there were more numbers! Doing it with 100 numbers was labourious enough! (That’s laborious for you americans). My initial thoughts were using modular arithmetic but I can’t find a way for that to work.
It did occur to me that this would be much easier if it were multi choice as it is easier to watch a numbers change in position and work out if it gets deleted than work out which ones stay in. (as I have done below with the 98th).
A few people have said 98 so i will show that to be incorrect.
on the first run through we discard 20 people leaving 80, person 100 is out.
on the second run through we discard 16 people leaving 64 people, and person 99 is gone.
At this point person 98 is a 4 and person 1 is a 5 (and discarded) we then dicard a further 11(thats 12 including person 1) (last person to discard is 61st on that list)
62nd is a 1 63rd is a 2 and 64 is a 3 making the new number 1 a 4
now as we just deleted 13 people we are left with 51 people at this point
we discard person 2, 7, 12, … up to and including the 47th on the list making the 51st now a 4 and the new number 1 a 5 (leaving 37 people)
Next round we get rid of 1, 6, 11, 16, .. 36 (8 people leaving 29) the 37th is a 1, the new 1st is a 2
Next round get rid of 4 , 9, 14, 19, 24 and 29 as 29 is the last on the list this is the original number 98 gone. It is irrelavant that 29 is not divisable by 5
2)
Just one possibility
the ones digit must be a 6
manjyomesando1 has explained this one fairly well already
any number can be written as 10x + y where y is a digit from 0 to 9 and x is any sized number
(10x + y)^2 = 100x^2 + 20xy + y^2
now the 100x^2 is irrelavant, it wont effect the last 2 digits
so 20xy + y^2 is what we need to look at
The ones digit is the last digit of y^2 which is 0, 1, 4, 5, 6, or 9
If y = 0 then 20xy + y^2 = 0 so the 10s digit is 0
if y = 1 then 20xy + y^2 = 20x + 1 for the 10s digit to be 7 is impossible because no matter what x is multiples of 20 will never have a tens digit of 7
if y = 2 then 20xy + y^2 = 40x + 4 again for the 10s digit to be 7 is impossible because no matter what x is multiples of 40 will never have a tens digit of 7
if y = 3 then 20xy + y^2 = 60x + 9 again for the 10s digit to be 7 is impossible because no matter what x is multiples of 60 will never have a tens digit of 7
if y = 4 then 20xy + y^2 = 80x + 9 again for the 10s digit to be 7 is impossible because no matter what x is multiples of 80 will never have a tens digit of 7
if y = 5 then 20xy + y^2 = 100x + 25 This always has a tens digit of 2 (not 7)
if y = 6 then 20xy + y^2 = 120x + 36 now if x = 2 this is 240 + 36 = 276 so 6 is possible
if y = 7 then 20xy + y^2 = 140x + 49 because y^2 is 49 we need to consider if the 10s digit of 140x will ever be a 3 (which it never will because multiples of 14 are even)
if y = 8 then 20xy + y^2 = 160x + 64 so we need to consider if the 10s digit of 160x will ever be a 1 but it never will because m,ultiples of 16 are even
if y = 9 then 20xy + y^2 = 120x + 81 consider if 12x ever ends in a 9, but never because 12x is always even
So we only found one possibility and that is a ones digit of 6
3)
Sketch at:
http://farm4.static.flickr.com/3032/2697142397_80b6bee5dd_o.jpg
Notice the red lines, the long red line is the radius of the dish and it bisects the 60degree angle of the equilateral triangle joining the centre of the spheres and then I have dropped a red line down perpendicular to the equilateral triangle
now from centre of dish to centre of a sphere lets call that d for distance
Cos(30) = r/d
d = r/cos(30)
d = 10/cos(30)
we add this to r (10cm) and we have the radius of the dish
radius of dish = 10 + 10/cos(30)
= 10 + 20/sqrt3
= 21.547cm (3dp)
NOTE: if it were 3r = 30cm then we would be able to fit a whole extra sphere between the 3 that are there but I think it’s obvious this would not fit. The answer must be greater than 20 and less than 30!
4)
someone posted this same question a few days ago, I think it was only 3 (no, it seems it was 4, manjyomesando1 caught one that I originally missed and have since added to my answer, I claim no credit for this extra one - I missed it)
At most you can pair 2 digits together unless it starts with 12, or 13
if it starts with 123 its still too high so at most you can pair 2 digits
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 so a minimum pairing has to be 99 but that’s impossible so we must have at least 2 pairings.
Start by pairing 8 and 9 and see how much short it is
1 + 2 + 3 + 4 + 5 + 6 + 7 + 89 = 117 so we are short by 27 by pairing 3 and 4 we get 27 more so
1 + 2 + 34 + 5 + 6 + 7 + 89 = 144 ******** (1)
Now start by pairing 78 and see how short we are:
1 + 2 + 3 + 4 + 5 + 6 + 78 + 9 = 108 this is 36 short
1 + 2 + 3 + 45 + 6 + 78 + 9 = 144 ********* (2)
12 + 34 + 5 + 6 + 78 + 9 = 144 ************ (3)
Now start by pairing 67 and see how short we are:
1 + 2 + 3 + 4 + 5 + 67 + 8 + 9 = 99 (short by 45 but no single pairing will fix this so we need to try 2 pairings)
12 + 3 + 45 + 67 + 8 + 9 = 144 works ************ (4)
Now is there anythign if we start by pairing 56
1 + 2 + 3 + 4 + 56 + 7 + 8 + 9 = 90, short by 54, would need 2 pairings but 12 + 34 + 56 + 7 + 8 + 9 is only 126 so no good.
1 + 2 + 34 + 5 + 6 + 7 + 89 = 144 ******** (1)
1 + 2 + 3 + 45 + 6 + 78 + 9 = 144 ********* (2)
12 + 34 + 5 + 6 + 78 + 9 = 144 ************ (3)
12 + 3 + 45 + 67 + 8 + 9 = 144 *********** (4)
are the only possibilities
