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How many eggs in the basket #43

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How many eggs in the basket

A women was carrying a basket of Eggs when a passer-by bumped her and she dropped the basket and all the eggs broke, Passer-by asked her number of eggs to pay her. Women replied, I don’t remember exactly , but I do recall that whether I divided the eggs by 2,3,4,5 or 6 there was always one egg left over. When I took the eggs out in groups of seven, I emptied the basket.

Can you tell, “How many minimum number of Eggs were there in the basket”?

Tried enough already?
See Solution
As in the previous problem we should find the least common multiple of 2,3,4,5 and 6 first, i.e. 60.
Now the the number of eggs can be 60+1, but its not completely divisible by 7.
So next possibility is 120+1, but its not completely divisible by 7 too.
So next possibility is 180+1, but its not completely divisible by 7 too.
So next possibility is 240+1, but its not completely divisible by 7 too.
So next possibility is 300+1, and its completely divisible by 7, so the minimum number of eggs in the basket should be 301.

Boy with Toffee #42

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Boy with Toffees

A boy has few toffees with him, then comes one of his friend and they decide to share them equally and they are able to divide them without any toffee remaining. Then one more friend comes and they divide the toffees in 3 equal parts, then one more friend comes, they are still able to divide the toffees in 4 equal parts, then comes one more friend, they are still able to divide them equally, similarly they are able to divide it equally in 5 and 6 equal parts but not when they are 7 and decide to divide equally they are left with one extra toffee.
Can you tell the minimum number of toffees boy had initially ?

Tried enough already?
See Solution
Since, They are able to divide the toffees equally between 2,3,4,5 and 6 boys so the number of toffees must be a multiple of all of these.
To get the minimum number we should first get the Least Common Multiple of 2,3,4,5 and 5, i.e. 2^2.3^1.5^1 = 60.
Now lets divide it by 7, the remainder is = 4, so it cant be the answer, So we should try the next possible answer i.e. 120, divide it by 7, remainder = 1, so minimum number of toffees boy can have is 120.

Average Age Puzzle #41

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Average Age Puzzle

Average age of a 10 members committee is same as it was 4 years ago, because an old member is replaced by a younger member. What is difference between the younger member and the old member age.
Or in other words, how younger he is from his counterpart?

Tried enough already?
See solution
Lets say sum of the 10 members 4 years ago was S, now it should have been S+40. assume old member’s age was x(now) and younger members age is y(now).
S + 40 -x +y = S=> x -y = 40.

9 digit number #46

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9 digit number puzzle

There is a 9 digit number. No digit are repeated and rightmost digit is divisible by 1 and right 2 digits is divisible by 2, right 3 digits is divisible by 3 and so on, finally the whole number is divisible by 9.

Can you find out the number?

See Solution
As the right two digits are divisible by 2, so rightmost digit should be even, at the same time right 5 digits are divisible by 5 so the rightmost digit can be 5 or 0, but as it should be even so it can only be 0.

Now, Right 4 digits are divisible by 4, so the rightmost two digits can be either 20, 40, 80 or 60.
To be divisible by 9 the digits of a number must sum to a multiple of 9. Adding the 10 digits together (0+1+2+…+8+9) gives 45 which is div by 9. However, this is a 9-digit number so we have to drop one from the ten. We can only take out 0 or 9 and still have the remaining digits sum to a multiple of 9. We’ve already shown that 0 is present so the 9 is dropped.
Right 3 digits are divisible by 3 so these 3 digits can be 120, 420, 720, 240, 540, 840, 180, 480, 780, 360.
Now to be divisible by 6 sum of the six digits should be divisible by 3 and it should also be even, which it is.

There is no rule for divisible by 7 so we just have to take digits such that they are divisible by 7, there can be many such numbers and one such number is 123567480.

Manage Your Energey Solution: Google codejam 2013 Round 1A

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Most of us are familiar with google codejam, those who don’t visit https://code.google.com/codejam for more information. This is the first problem from Online Round 1A 2013, top 1000 from this round will be eligible for next online round.

Problem Statement

You’ve got a very busy calendar today, full of important stuff to do. You worked hard to prepare and make sure all the activities don’t overlap. Now it’s morning, and you’re worried that despite all of your enthusiasm, you won’t have the energy to do all of this with full engagement.

You will have to manage your energy carefully. You start the day full of energy – E joulesof energy, to be precise. You know you can’t go below zero joules, or you will drop from exhaustion. You can spend any non-negative, integer number of joules on each activity (you can spend zero, if you feel lazy), and after each activity you will regain R joules of energy. No matter how lazy you are, however, you cannot have more than E joules of energy at any time; any extra energy you would regain past that point is wasted.

Now, some things (like solving Code Jam problems) are more important than others. For the ith activity, you have a value vi that expresses how important this activity is to you. The gain you get from each activity is the value of the activity, multiplied by the amount of energy you spent on the activity (in joules). You want to manage your energy so that your total gain will be as large as possible.

Note that you cannot reorder the activities in your calendar. You just have to manage your energy as well as you can with the calendar you have.
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Bullseye Solution: Google codejam 2013 Round 1A

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Most of us are familiar with google codejam, those who don’t visit https://code.google.com/codejam for more information. This is the first problem from Online Round 1A 2013, top 1000 from this round will be eligible for next online round.

Problem

Maria has been hired by the Ghastly Chemicals Junkies (GCJ) company to help them manufacture bullseyes. A bullseye consists of a number of concentric rings (rings that are centered at the same point), and it usually represents an archery target. GCJ is interested in manufacturing black-and-white bullseyes.

Maria starts with t millilitres of black paint, which she will use to draw rings of thickness 1cm (one centimetre). A ring of thickness 1cm is the space between two concentric circles whose radii differ by 1cm.

Maria draws the first black ring around a white circle of radius r cm. Then she repeats the following process for as long as she has enough paint to do so:

  1. Maria imagines a white ring of thickness 1cm around the last black ring.
  2. Then she draws a new black ring of thickness 1cm around that white ring.

Note that each “white ring” is simply the space between two black rings.

The area of a disk with radius 1cm is π cm2. One millilitre of paint is required to cover area π cm2. What is the maximum number of black rings that Maria can draw? Please note that:

  • Maria only draws complete rings. If the remaining paint is not enough to draw a complete black ring, she stops painting immediately.
  • There will always be enough paint to draw at least one black ring.

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Girl name in the car

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guess the girl name in the car

A girl was driving a car, a guy took a lift from her. He asked her name after he get down
Girl said :- my name is hidden in my car’s number, find if you can. Car number was [ WV733N ] Can you guess the girl’s name now?

See Answer
Answer:
It’s NEELUM.

Coins and sandwiches

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Three friends A, B and C went on a trip.
A didn’t carry any food with him but had 13 Coins. B had 6 sandwiches, and C had 7 sandwiches with him (Total 13 Sandwiches).

They decided to share the food equally and in return A would give away all 13 coins to B & C.

How should B & C distribute the money?


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Coins and sandwiches solution

Since they have to share the sandwiches equally, each of them will get 13/3 sandwiches.

Hence the sandwiched contributed by B and C to A.
B: 6 – 13/3 = 5/3
C:7 -13/3 = 8/3

Now A should give the money in the ratio of what he got from B and C, i.e.
5/3:8/3 :: 5:8

Hence B will get 5 and C will get 8 coins.

Guess the room number

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It was the first day of our second semester and we were looking for our hostel room numbers allotted, we saw that one of our friend was already allotted the room number last night and he challenged us to guess his new room number. We gave following answers

135, 780, 785 and 732.

He said, “That’s amazing”, “You’ve each guessed exactly one digit correctly and in its right place!”

Now can you guess what is his room number ?

See Guess the room number puzzle solution

Answer

Answer is 182

How?

explanation

The first digit must be 1 or 7. It can’t be 7 as only one digit is correct in each guess. So it’s 1.

The other digits in guess #1 (i.e. 3 and 5) must be incorrect. Therefore, from guess #3, the second digit is 8.

Guess #4 has a correct digit which must be its third, i.e. 2.

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