Expert: Josh - 10/26/2004

Question
                                                      abcd * 4 = dcba (all are different numbers)  

Answer
Hi Joe,

Here is what I propose. Before I answer this question, I think we should look at the metric system by way of an example. The aim here is to clarify the concepts and understand the relationship between a literal and the base 10 representation of a number. After that, we will work through the problem together. Then, I will give you one more practice question to try. Don't worry! The worked solutions are there. It's all good.

Part I: Number representations

Let us think about what a "string of digits" really mean? Consider the following question.

Question:  The unit digit of a two-digit number is 2 more than 4 times the tens digit. When the digits are reversed, the new number is 13 more than 3 times the original number. What is the original number?

Worked solutions: Let N be the unknown two-digit number.
Call the unknown digits which made up N "xy".
Here, x is the 'tens' digit, y is the 'unit' digit.
(If N=26, then, we have x=2, y=6. But so far, N is unknown)

Let's Interpret the information we were given:
(i) "unit digit of a two-digit is 2 more than 4 times the tens digit" means when we multiple x by 4, this is still 2 less than y. From this, we obtain 4x=y-2, rearranging gives:

4x+2=y (Equation 1)

(ii) when the digits are reversed, let's say R="yx", now we still use the same two unknown digits, but we simply write it backwards. So, y becomes the new 'tens' digit, x becomes the new 'unit' digit.

"When the digits are reversed, the new number (R) literally consists of "yx" is 13 more than 3 times the original number", we can think of this as

R=13+3N. (Equation 2)

But in order to solve for x and y, we must put x,y into equation 2, how do we do this?
We must ask ourselves what does it mean to be a 'unit' digit and a 'tens' digit.
.......................................................
Very simply, the metric system tells us that the 'tens' digit is to be multiplied by ten.
.......................................................
So, for the reversed number "yx":
    R=10y+x (Equation 3)
Similarly, for the original number "xy":
    N=10x+y (Equation 4)

We now have everything that we need. Substituting R and N into Equation 2, we have:

R=13+3N
10y+x=13+3(10x+y)
10y+x=13+30x+3y,
next, bringing everything to the left hand side, we get
-29x+7y=13 (Equation 5)
Everything boils down to two equations.
================
4x-y = -2  [1]
-29x+7y=13 [5]
================
Solving equation 1 & 5 simultaneously,
we can eliminate “y” to first find “x” by taking [5]+7*[1]

(4*7-29)x +(7-7)y= 13- 2*7, simplifying, we have
(28-29)x = 13-14,
-x=-1, so x=1.

Putting this value of x into [1] above gives us 4-y=-2, so y=6.

We should have N=16, R=61.
That's not it. Always check that the answer is indeed correct after all this effort.
Condition 1: is 1*4+2=6 ?   true!
Condition 2: is 16*3+13=61 ? yes! Since 48+13 =61
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Part II: Back to your original question
Can you now see how we can set up the problem using the same ideas. Instead of the units "tens" and "ones", we have the place values for "thousands","hundreds","tens" and "ones".

Translating the literal into an algebraic equation, in terms of base 10 (The reason we call it "base ten" is because each place value is given with respect to multiples of 1,10,100,1000 and so forth. Each unit may be obtained by raising ten to some exponent. For example, 1=10^0, 10=10^1, 100=10^2, 10=10^3.)

The EXPRESSION abcd * 4 = dcba becomes
4(a*1000+b*100+c*10+d)=(d*1000+c*100+b*10+a)
(4000-1)a+(400-10)b+(40-100)c+(4-1000)d=0 ...[#0]

[If you are unsure, pick a number. Let's say we pick 2345. Clearly, 2345 = 2*1000+3*100+4*10+5*1. So, the place values are a=2,b=3,c=4,d=5 respectively. This is what base-10 expansion is about.

We must satisfy the condition,
3999a+390b-60c-996d=0 ...[#1]

Since there are 4 unknowns, we have three degrees of freedom here. So, we have to be wise in finding the right combinations.

Suppose we choose a=2,d=8. Equation [#1] immediately reduces to
3999*2-996*8+390b-60c=0
30+390b-60c=0
60c-390b=30 ...[#2]
This is a good sign, why? Because it turns out that everything is in multiples of ten.

Note: This selection has been carefully chosen. Observe that 390b-60c must be even, since an even number multiplied by an odd/even number always yields an even number. So, we must somehow turn 3999a into an even number, to ensure that the equation works out. Otherwise, "a=1" would have been the obvious choice.
Consider this. If "a" was an odd integer, from [#1], the linear combination 390b-60c-996d must equal 3999a (an odd number), which is impossible. Regardless of what you choose for b,c and d, 390b-60c-996d remains even. There is no way that we can make 3999a and -(390b-60c-996d) equal.

Continuing from [#2],
60c-390b=30, select the most straightforward value for b,
Solving for c with b=1,
60c-390=30
c=420/60=7.

There you have it. a=2,b=1,c=7,d=8.

Check this. 2178*4 = 8712.

If you want extra practice, visit: www.geocities.com/joshcameron_ae/
go to the "question archive" and have a look at Question004. The answer is there too.

Let me know if you have any question.

Cheers.