Solution To The Puzzle For The Moderator

As the title of the puzzle suggests, the challenge was directed on me. But I never had the chance to try if I could handle it or not. Most probably not. Fortunately, two former students of mine stood up and save me from this puzzle quandary. Engr. Jeffrey Agbunag and Engr. Ramil Balisnomo belong to the last batch (and therefore youngest) of the CE students I handled for Reinforced Concrete Design. They now work in Manila as Structural Design Engineers and are planning to join us in Dubai in the near future.

For further clarity, I made minor alterations to their solution and added some explanations without changing the essence of their reasoning. So here it is.

A Puzzle For The Moderator

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?

The Solution:

by Jeffrey Agbunag and Ramil Balisnomo
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Let :
A = first digit of the first integer
B = second digit of the first integer
C = first digit of the second integer
D = second digit of the second integer
E = first digit of the third integer
F = second digit of the third integer
Then:
10A + B = first integer; 10B + A = the reverse of the first integer
10C + D = second integer; 10D + C = the reverse of the second integer
10E + F = third integer; 10F + E = the reverse of the third integer

From Condition 1:
(10A+B)+(10C +D) + 3(10E+F) = 242

From Condition 2:
(10B + A) – 0.50(10C + D) + 10F + E = 116

From Condition 3:
(10A + B) – (10D + C) = 10

The three equations when simplified and written in matrix form becomes:

The non-square coefficient matrix will show that no unique solution is possible because there are only three equations and there are six unknowns. We would therefore require three more independent equations. In order to produce the extra conditions, the duo came up with two remarkably original propositions:

The second paragraph of the puzzle talks of the third integer and its reverse having their halves and quarters as reverses. Thus,

1. The Agbunag Theorem:

“For the set of positive, single digit numbers, two - and only two - could be halved twice and still produce whole number results.” These are the numbers 4 and 8. Thus, they concluded that only the two-digit numbers 48 and 84 could satisfy the conditions of the second paragraph. However, they correctly pointed out that it could not be 84 because it would contravene the stipulations of Condition 1 (the sum would exceed 242 since 3 x 84 is already 252). Hence, the third digit is 48 and therefore

E = 4 and F = 8.

Adding these two equations to the matrix expression, we get

Still, there are more unknowns (6) than equations (5). So from where could we possibly acquire the 6th equation?

Enter Jeffrey’s and Ramil’s numerical acuity; and they come up with a simple but universal truth where most of us could only see a blank wall.

2. Balisnomo’s Theorem:

“If the difference between two 2-digit numbers is 10, their last digits must be the same.”

Thus, hidden in Condition 3, they pointed out that B must be equal to C by Balisnomo’s Theorem. In equation form:

B – C = 0

which when inserted into the matrix expression, we finally obtain an invertible coefficient matrix.

Explicitly, we obtain and write the inverse and perform the matrix multiplication, using Excel:

to obtain:

finally,

1st integer = 36
2nd integer = 62
3rd integer = 48

sana tama sagot namin. Hehehe.

God bless.

-Jeffrey and Ramil
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I have confirmed this with Carlo. TAMA ang sagot niyo! Congratulations and my admiration to both of you.

- icarus.
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Solutions for the other Problems:

A certain Marlon solve two out of the three problems which I posted as additions to Carlos lone puzzle. I am not sure if Marlon is a Louisian but he is one heck of a logical guy. I hope he will come forward and identify himself because he is surely a welcome addition to our rank. Here are the postings he made on the comment sections:

Marlon said...

Given : ABCDEx 4 = EDCBA
The answer is : 21978x 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.

2BCDEx 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

February 19, 2009 9:10 PM
Marlon said...

The secret is to notice that the answer has more letters (5 letters) than 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 <-- carry9END+ 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+101+R = 10R = 10-1R = 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 = N1+(1+E)+R = E+102+R = 10R = 10-2R = 8

So now we have: 1 11 <-- carry9END+ 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 <-- carry956D+ 1085 = 1065Y

Doing the same reasoning for D and Y and get the answer

9567+ 1085 = 10652

February 19, 2009 9:21 PM
ICARUS said...

Crisp and precise logic, nothing wasted, each line, each thrust and stroke goes straight for the jugular. Well done, Marlon!

I can tell you're a puzzle enthusiast yourself. Maybe you can send us some more interesting posers.

Thanks a lot.
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