Assuming that numbers continue to fill the 3×3 grids in the manner shown above, in which line and column will the number 600 be shown? You can calculate the answer, but you can also figure it out just by analyzing the pattern…
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. But once we found, we can remember the...
3rd Row 2nd Column
line 3 in column 200, or line 3 in column 2 of box 67
it was not easy to stop calculating the answer and try another approach to this puzzle…
finally I saw it: the sum of the figures in any number gives you the position in the grid.
i.e. 600 –> 6 + 0 + 0 = 6. Number 600 will be in 3rd row, 2nd column.
second column, last row
if we divide 10 by 9, 1 remains. 10 comes in 1st square
if we divide 20 by 9, 2 remains. 20 comes in 2nd square
if we divide 30 by 9, 3 remains. 30 comes in 3rd square
if we divide 600 by 9, 6 remains . so 600 comes in sixth square which is middle square of bottom row
yes
To get from one square in a grid to the same square in the next grid, we add 9. Which means that:
numbers in the 1st square -1 should be divisible by 9
numbers in the 2nd square -4 should be divisible by 9
numbers in the 3rd square -7 should be divisible by 9
numbers in the 4th square -2 should be divisible by 9
numbers in the 5th square -5 should be divisible by 9
numbers in the 6th square -8 should be divisible by 9
numbers in the 7th square -3 should be divisible by 9
numbers in the 8th square -6 should be divisible by 9
numbers in the 9th square -9 should be divisible by 9
Out of these, 600 is valid in the 8th square only, ie 3rd row, 2nd column
28 31 34
29 32 35
30 33 36
Here are the calculations (already figured out the pattern) :
1st square: 600-1/9=66.555555556 X
2nd square: 600-4=888888889 X
4th square: 600-2/9=66.444444444 X
5th square: 600-5/9=66.111111111 X
6th square: 600-8/9=65.777777778 X
7th square: 600-3/9=66.333333333 X
8th square: 600-6/9=66 yes
9th square: 600-9/9=65.666666667 X
So according to the calculations, the number 600 will be in column 1, line 3.
600 will be at the bottom middle of the 66th grid.
Middle Column
Bottom Row
lower middle
from box-to-box, each square increases by 9. The number “600″ will therefore appear in the square for which the current number “x,” when subtracted from 600, is evenly divisible by 9.
Further, 600/9 is 66-2/3. 2/3 of 9 is 6, so 600-6 is evenly divisible by 9.
Number MOD 9 = position corresponding to first grid
So, 600/9 = 66 remainder 6
i.e. 3rd line, 2nd column
The value 600 will be in box 67, column 2, line 3.
600 would be found line 3, column 2. In the same position as 6 in the first grid above. The number in the third line, third column will always be a multiple of 9. Finding the first multiple of nine above 600, I got 603; 9 * 67 = 603. So the grid that had 603 in the lower right corner would also include 600. Working backward in the pattern on the grid, 600 comes in lower center.
595 598 601
596 599 602
597 600 603
67th box 3rd row 2nd column
200th column
3rd line.
600/9 = 66,6 periodic.
66,6 times 3 = 200.
The periodic 6 is associated with all 2nd column/3rd line numbers so I guessed the line with that.
Simple but efficient.
Each cell in the grid follows the formula 9x+n, n being the value of the corresponding cell in the first grid.
So 9x+n=600 => n=600-9x, with n being [1,9]. We can deduce x to 66, so n=600-594=6.
So the cell holding the number 6 will eventually progress to be 600.
In number theory terms, the corresponding boxes in each grid represent the residue classes modulo 9. 600 is congruent to 6 mod 9, so it will appear in line 3, column 2. Incidentally, it appears in the 67th grid.
second row third column, i think…
It will be in the bottome line, middle column.
sum up all the digits of the number, divide the result by nine.
if the remainder is 1, number fall in 1st box (col 1, row 1)
if the remainder is 2, number fall in 2nd box (col 1, row 2)
and so on
if the remainder is 0, number fall in last box (col 3, row 3)
600 will fall in 2nd column, third line
uhmm, is it, 3rd line, 2nd row?
just guessing by looking at the pattern…
600 will appear at the same position as the number 6
600 = 594 + 6
594 is one of the multiples of 9 so will appear at the position of 9. So in the next box 600 will appear at the position of 6. ( Middle column, bottom row )
The numbers are sequential from top to bottom, and then from left to right.
Notice numbers divisile by 3 are in the bottom row. Numbers divisible by 9 are in the bottom row as well as the last column.
So 600 is divisible by 3, so we know it is in the bottom row in the 200th column.
66 x 9 = 594. So 594 is bottom row, last column of the 66th block. So 6 more numbers would be in the middle column, on the bottom row of the 67th block.
600 will be bottom center. Multiples of nine are always bottom right. 600 is not a multiple of nine, but 603 is (since the digits add up to nine.)
9 is at r3c3
9×11 is 99, so that is in r3c3
100 is next so that is in r1c1, so every 100 including 600 is in r1c1
I just did the next square. Oof. Too early in the morning for me…
Ah!
Bilbao, as soon as I posted my answer I knew there was something else that was going on with problem after I went back an re-read it.
The sum of the digits is the position in the block. Didn’t see that at all… Neat.
Great logic bilbao, its amazing how it works!
the alternative answer i suggested was to watch the multiples of 6 and 10.
all multiples of 6 will show on the bottom line, alternating between the middle column in a grid, and the first and third on the next.
but the great deal is to realize that all multiples of 10 will be in the same cell as its “starting number”
for instance: 5 will be in the same position as 50, 6 in the same position as 60. and the same thing will happen with the multiples of 100: 300 in the same position as 3, so 600 in the same position as 6.
jmart574 your reasoning is flawed as 599 is not divisible by 9. the reasoning you use above will wotk for 999 ( ie 1000 will be r1c1).
Easiest way of calculating this is: 600/9= 66.66667.
So we know its in 67th box. then 66*9= 594 – which is the last (biggest) number in the 66th box. so then we just count 6 more numbers with the ascending pattern shown in the 3 boxes above and you come to r3c2
Bilbao, I think your answer is the best. I don’t think I would’ve ever guessed something like that. Kudos to you!
That answer was pretty clever Bilbao, I try to analize your answer and I found this: the 6th square would be this:
46..49..52
47..50..53
48..51..54
when the figures on the numbers dont add a one digit number you have to apply the same algoritm recursevely: 49-> 4+9=13-> 1+3=4 I mean to figure that out was very cool man.
Great math bilbao !!!
All your comments are welcome.
I took the puzzle home in the back of my mind and unexpectedly an AHA! moment came to me. It is amazing how our brain works…
Thanks suineg for your additional explanation.