4chan archive /sci/ (index)
2012-09-05 11:10 5030573 Anonymous (tumblr_lz7ivs7kmd1qbxowoo1_500.jpg 500x375 33kB)
A brain teaser that took me entirely too long to figure out.... Take the number 1, 2, 3, and 4. They can be arranged into a multiplication problem as such: > 3 x 4 = 12 The numbers 1, 2, 3, 4, 5 can as well: >13 x 4 = 52 How can you arrange 1, 2, 3, 4, 5, 6 into this format, using each number once? >picture unrelated.

2 min later 5030578 Anonymous
53* 2 = 4 (mod 1)

7 min later 5030588 Anonymous
>>5030578 no 6

7 min later 5030589 Anonymous
>>5030573 I think I figured out a quick way to work this out, lemme draw it out.

10 min later 5030598 Anonymous
54 x 3 = 162

11 min later 5030599 Anonymous
54 x 3 = 162

11 min later 5030601 Anonymous
64/12= 5.3

23 min later 5030616 Anonymous (math puzzle.png 1920x1158 187kB)
Here you go, looks like some others beat me to it while I was drawing. I guess the question now is, is there some more general way to derive it and is there only one single solution? What I did was basically figured that the last value of both of the operands had to produce a last value that was still left in the pool (1*5 = 15, but there was no 5 left to use on the right side of the equation anymore). This allowed me to eliminate almost all of the candidates for the two last values on the operands and deduce possible last values for the right side.

35 min later 5030642 Anonymous
3: none 4: 3 x 4 = 12 5: 4 x 13 = 52 6: 3 x 54 = 162 7: none 8: 3 x 582 = 1746 6 x 453 = 2718 24 x 57 = 1368 34 x 52 = 1768 37 x 58 = 2146 58 x 64 = 3712 9: 4 x 1738 = 6952 4 x 1963 = 7852 12 x 483 = 5796 18 x 297 = 5346 27 x 198 = 5346 28 x 157 = 4396 39 x 186 = 7254 42 x 138 = 5796 48 x 159 = 7632

42 min later 5030653 Anonymous
I just used basic reasoning. The following will be very long in writing, but when you do it in your head it's very quick and obvious (except for the calculations, I had to write those down to not find myself repeating calculations I already did and not eliminating possibilities). First take some boundaries using numbers that aren't restricted to 1 to 6. 10 * 10 = 100, so having two digits * two digits will never give you two digits. 9 * 9 = 81, so having 1 digit * 1 digit will never give you 4 digits. 3 digits * 1 digit will give you 3 digits, which is 7 digits and out of range, so that's anything with 3 digits on the side with the operation is a goner. That leaves only 2 digits * 1 digit = 3 digits, since 6 - (2+1) = 3 Back to using the number restrictions of 1 to 6. The 2 digit number can never be in the 10s because 15 * 6 = 90. You can continue doing this to eliminate more candidates for the tens digit of the two digit number, but you don't have to because you can find the answer in another way. Now you have a format of (10x + y)*z, where x, y and z are numbers 1 to 6 and cannot equal one another, and this can be rewritten as 10x*z + y *z. 10x*z will never affect the ones digit for the three digit number, so all you have to focus on is y*z for the ones digit of the three digit number, which we will call q. Thus y*z = q, and they are all distinct integers in 1 to 6 range. 1 * z = z, and only 1 * 1 = 1, which is a contradiction, so 1 will never be in these y, z, q slots. A: 2 * 3 = 6 B: 3 * 4 = 2 ...are your ONLY options left for y, z and q.

43 min later 5030656 Anonymous
>>5030653 From the above reasoning, we know x will never be 1. Thus 1 must end up on the hundreds or tens digit of the 3 digit number. 3 is either in the two digit number or in the one digit number, and cannot be elsewhere. 2 cannot be x because it must be either y, z or q. Therefore, x must be either 4, 5 or 6. If x = 6, then you must go with option B or else you reach a contradiction, and if x = 4 you must go with option A. Let's try x = 4. 42 * 3 = 126, which fails. 43 * 2 = 86 also fails. If x = 6, 63 * 4 = 252 fails, and 64 * 3 = 192 also fails. Thus x = 5. If we try A, then 52 * 3 = 156, and 53*2 = 106, and these both fail. If we try B, then 53 * 4 = 212 fails, but 54 * 3= 162 matches all the criteria for the answer. The answer is found.

53 min later 5030668 Anonymous
>>5030656 As I said, when it's formalized, it's very clunky and ridiculous because you have to communicate to others your mental models using stuff that's mentally unnecessary like 'y*z = q', but it's straightforward to do it intuitively. Basically, find the structure of 2 digits * 1 digit = 3 digit (this is intuitive if you know your multiplication well) and you know y*z = q only has a few options. You know 1 cannot be in either of these slots, nor can it be in x, and you know 2 and 3 cannot be in x. You also know x = 4 and 6 bind you to certain options for y,z,q. After that it's really easy to do.

1 hours later 5030682 Anonymous
4+5+6-3=12 Masterrace reporting

1 hours later 5030727 Anonymous
>>5030682 addition/subtraction?

1.614 0.057