Goldman Sachs

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# `Goldman Sachs` `Software Engineer Intern` Interview Question

`"Suppose you had eight identical balls. One of them is slightly heavier and you are given a balance scale . What's the fewest number of times you have to use the scale to find the heavier ball?"`

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Part of a `Software Engineer Intern` Interview Review - one of `1,564` `Goldman Sachs` Interview Reviews

`Goldman Sachs`

Answers & Comments

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of 44votes

Two. Split into three groups of three, three, and two. weigh the two groups of three against each other. If equal, weigh the group of two to find the heavier. If one group of three is heavier pick two of the three and compare them to find the heaviest.

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of 5votes

Brian - this would be correct if you in fact were using a weighing scale, and not a balance scale. The ability to weigh one group against another with a balance scale allows Marty's answer to be a correct answer.

Although - the question as worded provides a loophole. If it had been worded as "What's the fewest number of times you have to use the scale to CONSISTENTLY find the heavier ball", then Marty's answer would be the only correct answer.

However, it is possible that you could get lucky and find the heavier ball in the first comparison. Therefore, the answer to the question as stated, is ONE.

Although - the question as worded provides a loophole. If it had been worded as "What's the fewest number of times you have to use the scale to CONSISTENTLY find the heavier ball", then Marty's answer would be the only correct answer.

However, it is possible that you could get lucky and find the heavier ball in the first comparison. Therefore, the answer to the question as stated, is ONE.

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2

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of 3votes

This question is from the book "How to move Mt Fuji".... Marty has already got the right answer.

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of 5votes

Actually Bill, by your interpretation of the question the answer is zero, because you could just pick a ball at random. If you get lucky, then you've found the heaviest ball without using the scale at all, thus the least possible amount of times using the scale would be zero.

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of 2votes

The answer is 2, as @Marty mentioned. cuz its the worst case scenario which u have to consider, otherwise as @woctaog mentioned it can be zero, u just got lucky picking the first ball....

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of 6votes

None- weigh them in your hands.

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of 6votes

Assuming that the balls cannot be discerned by physical touch, the answer is 3.

You first divide the balls in two groups of 4, weigh, and discard the lighter pile. You do the same with the 4 remaining, dividing into two groups of 2, weighing, and discarding the lighter pile. Then you weigh the two remaining balls, and the heavier one is evident.

You first divide the balls in two groups of 4, weigh, and discard the lighter pile. You do the same with the 4 remaining, dividing into two groups of 2, weighing, and discarding the lighter pile. Then you weigh the two remaining balls, and the heavier one is evident.

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6

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of 7votes

2

3a+3b+2 = 8

if wt(3a)==wt(3b) then compare the remaining 2 to find the heaviest

if wt(3a) !== wt(3b) then

ignore group of 2

discard lighter group of 3

divide the remaining group of 3 into 2+1

weigh those 2

If == the remaing 1 is the heaviest

if !== the heaviest will be on the scale

3a+3b+2 = 8

if wt(3a)==wt(3b) then compare the remaining 2 to find the heaviest

if wt(3a) !== wt(3b) then

ignore group of 2

discard lighter group of 3

divide the remaining group of 3 into 2+1

weigh those 2

If == the remaing 1 is the heaviest

if !== the heaviest will be on the scale

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of 3votes

With the systematic approach, the answer is 3.

But, if you randomly choose 2 balls and weigh them, and by coincidence one of these two is the heavier ball, then the fewest number of times you'd have to use the scale is 1.

Although the real question is: are the balls truly identical if one is heavier than the rest?

But, if you randomly choose 2 balls and weigh them, and by coincidence one of these two is the heavier ball, then the fewest number of times you'd have to use the scale is 1.

Although the real question is: are the balls truly identical if one is heavier than the rest?

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of 3votes

just once. Say you are lucky and pick the heavy ball. One use of the scale will reveal your lucky choice

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of 2votes

so once, or the creative answer zero if you allow for weighing by hand

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of 0votes

Without judging by hand: Put 4 balls on one side, and 4 on the other. Take the heavier group and divide again, put 2 balls on one side, and 2 on the other. Take the 2 that were heavier, and put one on each side. You've now found the heaviest ball. This is using the scale 3 times, and will always find the right ball.

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of 1vote

None. They are identical. None is heavier.

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of 4votes

2 weighings to find the slightly heavier ball.

Step 1.

compare 2 groups of three balls.

Case 1.

if they are both equal in weight, compare the last 2 balls - one will be heavier.

case 2.

If either group of 3 balls is heavier, take 2 balls from the heavier side.

compare 1 ball against the 2nd from the heavy group

result 1. if one ball is heavier than the other, you have found the slightly heavier ball.

result 2. if both balls are equal weight, the 3rd ball is the slightly heavier ball.

Easy Shmeezi

Step 1.

compare 2 groups of three balls.

Case 1.

if they are both equal in weight, compare the last 2 balls - one will be heavier.

case 2.

If either group of 3 balls is heavier, take 2 balls from the heavier side.

compare 1 ball against the 2nd from the heavy group

result 1. if one ball is heavier than the other, you have found the slightly heavier ball.

result 2. if both balls are equal weight, the 3rd ball is the slightly heavier ball.

Easy Shmeezi

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of 0votes

Fewest - get lucky and pick the heaviest one. But wait! How would you know it is the heaviest one by just weighing one ball? Your logic is flawed.

Two groups of four. Split heavier one, weigh. Split heavier one, weigh. 3 times.

Two groups of four. Split heavier one, weigh. Split heavier one, weigh. 3 times.

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of 2votes

i think its 3. i would take it like this

OOOO OOOO then

OO OO

then

OO

problem solved. i do this everyday. bye. praise be to allah. thats it.

OOOO OOOO then

OO OO

then

OO

problem solved. i do this everyday. bye. praise be to allah. thats it.

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of 1vote

It's 2. Period.

If you can't figure it out look it up online or in "How Would You Move Mount Fuji" (like somebody else said). This is one of the most basic brainteasers you could be asked in an interview.

If you can't figure it out look it up online or in "How Would You Move Mount Fuji" (like somebody else said). This is one of the most basic brainteasers you could be asked in an interview.

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3

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of 3votes

The answer is 2.

1) Divide the balls into 3 groups. 2 piles with 3 balls each, 1 pile with 2 balls.

2) Weigh the 2 piles of 3 balls. If both piles are the same weight, discard all 6 and weigh the last 2 to find the heavier one.

3) If 1 pile of 3 is heavier than the other, discard the lighter pile and the pile of 2 balls. Weigh 2 of the remaining 3 balls from the heavier pile. If both of the weighed balls are equal, the last ball is the heavier one.

1) Divide the balls into 3 groups. 2 piles with 3 balls each, 1 pile with 2 balls.

2) Weigh the 2 piles of 3 balls. If both piles are the same weight, discard all 6 and weigh the last 2 to find the heavier one.

3) If 1 pile of 3 is heavier than the other, discard the lighter pile and the pile of 2 balls. Weigh 2 of the remaining 3 balls from the heavier pile. If both of the weighed balls are equal, the last ball is the heavier one.

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of 1vote

2=if all the balls are identical and you pick up the first...weigh it and the second one is lighter or heavier then you've found the heavier ball in the least amount of attempts.

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of 1vote

1=if all the balls are identical and you pick up the first...balance it and the second one is lighter or heavier then you've found the heavier ball in the least amount of attempts.

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of 0votes

Amy is 100% correct for the following reason:

everyone (except Amy) is solving the theoretical problem. The practical side of the problem - notwithstanding jimwilliams57's brilliant observation that if one weighs more than the others IT IS NOT IDENTICAL (would have loved to see the interviewer's face on that one) - in order to 'weigh' them on a scale, one has to pick them up, therefore, you will immediately detect the heavier one without weighing: pick-up three and three ... no difference, no need to weight. Pick-up the remaining two to determine the heavier one.

Steve

everyone (except Amy) is solving the theoretical problem. The practical side of the problem - notwithstanding jimwilliams57's brilliant observation that if one weighs more than the others IT IS NOT IDENTICAL (would have loved to see the interviewer's face on that one) - in order to 'weigh' them on a scale, one has to pick them up, therefore, you will immediately detect the heavier one without weighing: pick-up three and three ... no difference, no need to weight. Pick-up the remaining two to determine the heavier one.

Steve

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of 0votes

First off, take yourself through the process visually and forget square roots, that doesnt apply here and here is why: The question is the Minimum, not the MAXIMUM. BTW, the max would be 7 ( 8-1); you are comparing 2 objects, so 1 ball is eliminated automatically in the first step. Anyway, you have a fulcrom of which you are placing 2 of 8 objects on each end. If by chance you pick the slightly heavier object as one of the two balls, you have in fact, found the slightly heavier one in the first round... btw dont be a smartass with your interviewer, he is looking for smarts not smarmy;)

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of 0votes

Respectfully, the folks who are answering "3" are mathematically modeling the nature of the balance incorrectly. Performing a measurement on a balance scale is not binary. It is trinary. Each measurement gives you one of three responses: The left is heavier, the right is heavier, or they are equal. So while you do need three binary bits to specify a number from one to eight, you need only two TRINARY-DIGITS

Formally, you want the smallest value of n such that 3^n >= 8. The answer is 2. Note that you could add a ninth ball, and still, you'd only need to make two measurements.

Of course, the smarty pants answer would be one. Just pick two balls at random and be lucky to have chosen the heavy one. But you're not guaranteed to be able to do it in just one measurement.

Formally, you want the smallest value of n such that 3^n >= 8. The answer is 2. Note that you could add a ninth ball, and still, you'd only need to make two measurements.

Of course, the smarty pants answer would be one. Just pick two balls at random and be lucky to have chosen the heavy one. But you're not guaranteed to be able to do it in just one measurement.

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of 0votes

English isn't my mother tongue... What is a balance scale? If looking up on Google, I find some devices with two bowls on a bar bearing in the center. Hence, the answer is once (if I'm luck enough to select the heavier ball in teh first measurement) If a balance scale allows to measure only one ball at a time, then it would take two measurements, unless you'd have more information on the weight, which is not listed here, hence doesn't exist in the context of the question^.

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of 0votes

3 times. Not having looked at the other comments, hopefully, I am the 26th to get this right.

Put the balls 4 and 4 on the scale, Take the heavier side and place those balls 2 and 2 on the scale. Take the heavier side and place them 1 and 1 giving the heaviest ball.

Put the balls 4 and 4 on the scale, Take the heavier side and place those balls 2 and 2 on the scale. Take the heavier side and place them 1 and 1 giving the heaviest ball.

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of 0votes

OK, now I read the comments and see that the people, like the question are divided into to groups, systematic approach people that say 3 (like I did) and analytic people that say 2. It takes a systematic person (me) a minute to get the answer. I'm guessing it took the analytic 5 minutes just to interpret all the ramifications of the question, i.e. they aren't idenitical if..., do it by hand..., get lucky.

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of 0votes

minimum is 1 (if lucky - 25% chance by picking 2 balls at random) & max is 2 (using most efficientl process to absolutely determine without luck - 3/3/2 scenario)

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of 1vote

While Symantec was busy weighing my balls I took a job with NetApp.... They need to focus on hiring good, capable security engineers, not weighing their balls.

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of 0votes

The point of these interview questions is to both check your logical brain function and to hear how you think. Most of you are just posting jerk off answers trying to be funny, or you are really dumb. These answer get you nowhere with me in an interview. Think out loud, go down the wrong path back track try another logic path, find the answer. None of this "0 if you use your hands". That is fine if you are interviewing for a job in advertising where creativity is desired, nobody wants you writing code like an 8 year old.

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of 0votes

You have 12 balls, equally big, equally heavy - except for one, which is a little heavier.

How would you identify the heavier ball if you could use a pair of balance scales only twice?

How would you identify the heavier ball if you could use a pair of balance scales only twice?

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of 0votes

The problem is based on Binary Search. Split the balls into groups of 4 each. Choose the heavier group. Continue till you get the heavier ball. This can be done in log(8) (base 2) operations, that is, 3.

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votes

BrianonJul 29, 2009