Carlo sent in an amusing challenge problem and calls it “a puzzle for the moderator”. Obviously, he’s got enough of my brain-teasers and this time he wants to be the one to bewilder; and not the bewildered. It’s the young lion of the pride, newly emerged from being a cub, playfully testing his power against the aging alpha male. It’s the promising pupil gauging his washed-up teacher, the growing son measuring his fading father. And this is all very healthy. As in everything, the new will shade the old - in due time. I am, of course, much obliged to solve it but before I do, I will give everyone the chance to test their mettle against it. So here it goes…
A Puzzle For The Moderator
by Engr. Carlo E. Chan
A Puzzle For The Moderator
by Engr. Carlo E. Chan
There are three positive integers, each with two non-repeating-digits. The sum of the first and the second integers when added to thrice the third integer results to 242. When half of the second integer is subtracted from a number which is the reverse of the first integer, and the difference is added to a number which is the reverse of the third integer, we obtain 116. If we reverse the second integer and deduct the resulting number from the first integer, we get a difference of 10.
Amusingly, the third integer has other amazing properties. When the number and its reverse is divided by two, the halves are reverses of each other. The halves when further halved are also reverses. Can you give me the integers?
Amusingly, the third integer has other amazing properties. When the number and its reverse is divided by two, the halves are reverses of each other. The halves when further halved are also reverses. Can you give me the integers?
<<<<<<<<<<<<<<<<<>>>>>>>>>>>>>>>>>
And for your further amusement, here are three other challenging number puzzles in addition to Carlo's.
Each letter in the three problems that follow represents a unique numerical digit (for that particular problem). Figure out the number digits represented by each letter so that no inconsistency or repetition arises when the indicated operations are performed.
As usual, send your solutions with the accompanying explanations on how the answers were arrived at.
Happy weekend, everyone!
<<<<<<<<<<<<< >>>>>>>>>>>>>
Given:
ReplyDeleteABCDE
x 4
= EDCBA
The answer is:
21978
x 4
= 87912
But rather than just give you the answer, here's how I figured it out. First, it is obvious that A must be an even number, because we are multiplying by 4 (an even number). The last digit will therefore be even. It can't be 0, because that would make ABCDE a four-digit number. It can't be more than 2, because that would result in a six-digit answer. So A = 2.
2BCDE
x 4
= EDCB2
So what can E be? The choices are E = {3, 8} because 3 x 4 = 12 and 8 x 4 = 32. But a value of 3 doesn't work in the result (3????) because it is too small.
2BCD8
x 4
= 8DCB2
Since the final number is 8 and we have 2 x 4, that means there is no carry from the prior multiplication (4 x B + carry). So B can't be anything higher than 1, possibly 0.
Looking at the other side of the equation, we have 4D + 3 = (a number ending in 0 or 1). In other words, 4D must end in 7 or 8. Obviously only 8 works, because 4 is an even number. Working forward again, that means B = 1.
21CD8
x 4
= 8DC12
So what values of 4D result in a number ending 8? 4 x 2 = 8, 4 x 7 = 28. Now 2 is already taken and the problem said the digits were unique. So D = 7.
21C78
x 4
= 87C12
Finally, we have a carry of 3 (from 28 + 3 = 31). And when we calculate 4C + 3 it must also result in a carry of 3 and a last digit of C. In other words:
4C + 3 = 30 + C
This is easy to solve:
3C = 27
C = 9
Thus the final answer is:
21978
x 4
= 87912
The secret is to notice that the answer has more letters (5 letters)
ReplyDeletethan the question (4 letters).
That M at the beginning of money is a carry from the thousands place,
so M = 1. Now we have:
SEND
+ 1ORE
-------
= 1ONEY
Now, in the thousands place there is a 1, so the only value for S that
could cause a carry is S = 9 and that means O = 10. Now we have:
9END
+ 10RE
-------
= 10NEY
Now look at the hundreds place. If there were no carry from the tens
place, E and N would be the same because E+0 = N, but E and N can't be
the same, so there must be a carry from the tens place. Now we have:
1 1 <-- carry
9END
+ 10RE
-------
= 10NEY
Now the equation for the hundreds place is 1+E+0 = N or just 1+E = N.
In the tens place we can have N+R = E+10 if there is no carry from the
ones place, or we can have 1+N+R = E+10 if there is.
First test: no carry from the ones place:
N+R = E+10 and 1+E = N
(1+E)+R = E+10
1+R = 10
R = 10-1
R = 9
But S = 9, so R cannot = 9. That means there is a carry from the ones
place and we get:
1+N+R = E+10 and 1+E = N
1+(1+E)+R = E+10
2+R = 10
R = 10-2
R = 8
So now we have:
1 11 <-- carry
9END
+ 108E
-------
= 10NEY
N cannot be 0 or 1 because 0 and 1 are taken.
N cannot be 2 because 1+2+8 = 11 and then E would equal 1, but it
cannot equal 1 because 1 is taken.
N could equal 3,4,5,6 or 7 but it cannot equal 8 or 9 because 8 and 9
are taken (and E must be 2,3,4,5 or 6 because it is 1 smaller than N).
If E were 2, then for the ones place to carry D would have to be 8 or
9, and both 8 and 9 are taken, so E cannot be 2 (and N cannot be 3).
If E were 3, then for the ones place to carry D would have to be 7,8,
or 9, but D cannot be 7 because then Y would be 0, which is taken, so
E cannot be 3 (and N cannot be 4).
If E were 4, then for the ones place to carry D would have to be
6,7,8, or 9. D cannot be 6 because Y would be 0, but D cannot be 7
because Y would be 1, and 0 and 1 are both taken, so E cannot be 4
(and N cannot be 5).
The only two possibilities for E now are 5 and 6.
If E were 6, then N would be 7 and D would have to be 4 (which would
make Y = 0), 5 (which would make Y = 1), 6 (which is taken by E), or
7 (which is taken by N). There are no solutions for E = 6, so E must
be 5.
So now we have:
1 11 <-- carry
956D
+ 1085
-------
= 1065Y
Doing the same reasoning for D and Y and get the answer.
9567
+ 1085
-------
= 10652
Crisp and precise logic, nothing wasted, each line, each thrust and stroke goes straight for the jugular. Well done, Marlon!
ReplyDeleteI can tell you're a puzzle enthusiast yourself. Maybe you can send us some more interesting posers.
Thanks a lot.
given the three non-repeating integers are ab, cd, ef, the three equations should be:
ReplyDeleteab + cd + 3ef = 242
ba - .5cd + fe = 116
ab - dc = 10
by commutative property ab=ba, and ef=fe
the third equation doesn't have the third term, ef, so my goal would be to get rid of ef. i can do that by multiplying the second equation by -3, then simplifying both equations to terms ab and cd only:
ab + cd + 3ef = 242
-3ab + 1.5cd - 3ef = -348
combine the terms vertically and the resulting equation should be: -2ab + 2.5cd = -106, then take the third equation and multiply that whole equation by 2:
-2ab + 2.5cd = -106
2ab + 2cd = 20
again combine the terms vertically and you get
.5cd = -86, isolate cd and you get:
cd = -172
substitute cd in the third (original) equation and solve for ab:
ab = -162
substitute ab and cd in the first (original) equation and isolate ef:
ef = 192
Whew! Give me a whiteboard and a marker anytime icarus! Writing out your solution in words is something else. I hope I'm right...
ricky
Noooo! When you say "reverse" do you mean the opposite of the integers?
ReplyDeletericky
hi ricky! I havent started solving it myself but it seems you missed one of the constraints - the numbers are just of two digits and positive. And yes, reverse means the numbers in the tens and ones are interchanged. Am I right Carlo?
ReplyDeleteYup that's right, you can think of it as an old digits problem back in college. But be wary, there's more to the surface than it appears and you might need some couple of tricks beyond the classic solution to get it done. Good luck!
ReplyDelete