Chessboard squares

Logic Level 1

The numbers 1 to 64 are written in order on the chessboard above. What is the sum of the numbers in the yellow squares minus the sum of the numbers in the white squares?

64 0 -64

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13 solutions

Jeremy Bansil
Feb 6, 2015

Let's say that we add diagonally opposite squares... 64 + 1 = 65 64+1=65 , 63 + 2 = 65 63+2=65 , 62 + 3 = 65 62+3=65 , 61 + 4 = 65 61+4=65 , and so on and so forth... All diagonally opposite squares have the same sum. And since there's an equal number of white and yellow squares, we can say that the answer is 0 \boxed{0} .

Nice. There are many ways of showing the sum is 0. Let's see how many others the community can come up with.

Calvin Lin Staff - 6 years, 4 months ago

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Since the number of squares on the chessboard is even this could happen. Isn't it? The problem would be more interesting if set for 7 × 7 7\times 7 chessboard!

Snehal Shekatkar - 6 years, 4 months ago

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I believe the intention is for this to be an even sided square, and then for people to figure out ways why this is equal to 0. There are many ways to do so. Mine is "any 2 by 2 square clearly has a sum of 0", hence we are done.

Asking if for a 7 by 7 square would make it less obvious, and it then becomes "do this somewhat tedious calculation". Of course, there are ways of speeding it up, by making certain observations. Perhaps, if you post a question for the 101 by 101 square? That would discourage people from finding 1 2 + 3 4 + . . . 1 -2 + 3 - 4 + ... , and hopefully inspire them to think creatively.

Calvin Lin Staff - 6 years, 4 months ago

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@Calvin Lin Right. Perhaps I can post that. However my only point was that the broken symmetry of the problem (numbers are increasing only in the right direction) makes the answer non-zero when square is odd.

Snehal Shekatkar - 6 years, 4 months ago

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@Snehal Shekatkar Yes, that would make a nice question. We can use a similar approach, to pair up the smaller 2n by 2n board, leaving only the edge, which we can easily sum. That will make a nice harder version of this problem.

Chung Kevin - 6 years, 4 months ago

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@Chung Kevin The odd number square matrix can be easily done with the help of AP therefore it wouldn't make a tough problem.

Prathmesh Pathwar - 6 years, 4 months ago

@Snehal Shekatkar Right. The odd case also has a nice generalization.

Calvin Lin Staff - 6 years, 4 months ago

the numbers in the 1st row of the yellow squares are 2,4,6,8.......which are in A.P then

then Tn = 2n then sum = S(Tn)

 sum = n * (n + 1)   but n=4  then

 sum = 4 * (4 + 1)

 sum = 20

the numbers in the 1st row of the yellow squares are 9,11,13,15.......which are in A.P then

Tn = 2n + 7

sum = S(Tn)

sum = n * (n + 1) + 7n n =4

sum = 4 * 5 + 7 * 4

sum = 48

the numbers in the 1st row of the yellow squares are 18,20,22,24.......which are in A.P then

Tn = 2n + 16

sum = S(Tn)

sum = n * (n + 1) + 16n n =4

sum = 4 * 5 + 16 * 4

sum = 84

the numbers in the 1st row of the yellow squares are 25,27,29,31.......which are in A.P then

Tn = 2n + 23

sum = S(Tn)

sum = n * (n + 1) + 23n n =4

sum = 4 * 5 + 23 * 4

sum = 112

the numbers in the 1st row of the yellow squares are 34,36,38,40.......which are in A.P then

Tn = 2n + 32

sum = S(Tn)

sum = n * (n + 1) + 32n n =4

sum = 4 * 5 + 32 * 4

sum = 148

the numbers in the 1st row of the yellow squares are 41,43,45,47.......which are in A.P then

Tn = 2n + 39

sum = S(Tn)

sum = n * (n + 1) + 39n n =4

sum = 4 * 5 + 39 * 4

sum = 176

the numbers in the 1st row of the yellow squares are 50,52,54,56.......which are in A.P then

Tn = 2n + 48

sum = S(Tn)

sum = n * (n + 1) + 48n n =4

sum = 4 * 5 + 48 * 4

sum = 212

the numbers in the 1st row of the yellow squares are 57,59,61,63.......which are in A.P then

Tn = 2n + 55

sum = S(Tn)

sum = n * (n + 1) + 55n n =4

sum = 4 * 5 + 55 * 4

sum = 240

the total sum of yellow squares = 20 + 48 + 84 + 112 + 148 + 176 + 212 + 240 = 1040

yellow squares sum = 1040

then

sum of white squares = sum of the total chess board - yellow squares sum

sum of white squares = ( 64 * 65 )/2 - 1040

sum of white squares = 2080 - 1040

sum of white squares = 1040

i.e

the sum of white squares = the sum of yellow squares then

the sum of white squares - the sum of yellow squares = 1040 - 1040 = 0

Ajey Devadiga - 5 years, 1 month ago

even i thought exactly the same way at first instance.....upvoted

Rajat Pathak - 5 years, 1 month ago

Take any small 2x2 square and do as indicated the result is zero, this pattern persist in the whole board. So dividing our 8x8 board into 16. 2x2 miniboard we come out with zero too.

Great observation!

Calvin Lin Staff - 6 years, 4 months ago

That is not completely true though. Only because the number of squres is even, this could happen. Try it!

Snehal Shekatkar - 6 years, 4 months ago

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You are correct I assumed otherwise the number of yellows and whites will be unequal.

Mariano PerezdelaCruz - 6 years, 4 months ago

A pattern of two by two squares is ALWAYS even.

Guy Josserand - 2 years, 10 months ago

I saw it too but more obvious is the difference between 2 consecutive integers is is 1.

Roman Frago - 6 years, 4 months ago
Ethan Godden
Feb 6, 2015

Of course, I did it a harder way by noticing a column equals 32 or -32 depending on the colour that starts. I only had to add up one column because each square as you go down is 8 more than the previous square. That means if you start from the top, the 2nd colour will have 8 more per square. of that colour. Since there are four of each colour in a column, the 2nd colour in each column will have 8x4 value more than the colour that starts.

Great observation!

Calvin Lin Staff - 6 years, 4 months ago

oh wow. I was not aware that the columns summed up so nicely. Thanks for sharing.

Chung Kevin - 6 years, 4 months ago
Brock Brown
Feb 6, 2015

Python:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
white = 0
yellow = 0
current = 'y'
for n in xrange(1,65):
    if current == 'w':
        white += n
        current = 'y'
    else:
        yellow += n
        current = 'w'
    if n % 8 == 0:
        if current == 'w':
            current = 'y'
        else:
            current = 'w'
print "Answer:", yellow - white

easy, we gonna subtract the yellow square to the white square

in the first row :

2 1 = 1 , 4 3 = 1 , 6 5 = 1 , 8 7 = 1 2-1=1, 4-3=1, 6-5=1, 8-7=1

for the second row :

9 10 = 1 , 11 12 = 1 , 13 14 = 1 , 15 16 = 1 9-10=-1, 11-12=-1, 13-14=-1, 15-16=-1

so, we got 4 in the the even-numbered row and -4 in the odds-numbered row

and since there's equal even-numbered and odds-numbered row, so the answer is 0 \boxed{0}

Krishna Garg
Jul 14, 2015

WE Added fist two rows white squares and also the yellow squares and subtracting the sums,it is ZERO.Same will apply for other rows as well. Therefore answer is ZERO.

It is a 8 * 8 Square Matrix, , We can think of it as Sum of sixteen 2 * 2 square matrices,, Now in each 2 * 2 matrix at top row yellow wins with 1 point, and at bottom row white wins with 1 points,,, Ultimately no one wins, the game ties.

Nice way of thinking about it

Chung Kevin - 6 years, 4 months ago
Mxjd Ultimate
Jun 5, 2021

Lets add the numbers seperaratly so hat it would add to 65. Since there is the same amount of white and black squares, the answer is 0.

Caillen Tu
Jun 14, 2020

Count the yellow squares. The answer should be 32. Then count the white squares. The answer also should be 32.

Then do the simple math. The answer is 0.

Benjamin Zandstra
Feb 15, 2019

I just add them all up and then subtracted.

Venkatachalam J
May 9, 2016

Sum(yellow squares-white squares) of first row from top is =4, Sum(yellow squares-white squares) of second row from top is=-4, Sum(yellow squares-white squares) of 1st row and 2nd row fro top=4-4=0.

Similarly, Sum(yellow squares-white squares) of 3rd and 4th row from top=0,Sum(yellow squares-white squares) of 5th and 6th row from top=0, & Sum(yellow squares-white squares) of 7th and 8th row from top=0. Therefore the answer is '0' (zero)

This is much easier than the column totals

Zahid Hussain - 1 year, 10 months ago
Luis Chavez
Apr 6, 2016

Half are white, half are brown. There are 32 of each. Thus subtracting them gives you zero.

Syed Baqir
Jul 29, 2015

Using Summation for Arithmetic series:

Sum of White boxes= 1/2 * 32 * (1 + 64) = 1040

Sum of Yellow boxes = 1/2 * 32 * ( 2 + 63) = 1040

Hence 1040 - 1040 = 0 (ANS)

I would say this is the most logical yet easiest to understand proof.

Zahid Hussain - 1 year, 10 months ago

My apologies. This "proof" is actually flawed because the numbers in both the series do not follow a regular pattern. e.g. number in first series are 1 3 5 7 10 12 ... which actually is not an arehtimetic series

Zahid Hussain - 1 year, 10 months ago

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