. . .

 Subj:..Find A Four-Digit Integer II         From: The Puzzlet Page        on 7/9/2008 (S652) Drawing from Valdosta State University...
Source: http://www.puzzlet.co.uk/Puzzlets/Puzzlet_001.html

Find a 4-digit integer has the following properties:
The sum of its first and third digits equals
the sum of the other digits.
The product of its first and last digits is ten times
greater than the product of the remaining digits.
The sum of the first two digits is only half
the sum of the last two digits.

What is the integer?  Is there more than one answer?

¤»¥«¤»§«¤»¥«¤»§«¤»¥«¤»§«¤»¥«¤»§«¤»¥«¤»§«¤»¥«¤»§«¤

Hint: Since there are four unknowns and only three
equations, multiple solutions should be expected.

.
.
 Finger pointing down from darrell94590 on 1/2/2006
.

 Drawing from Ripleys-Believe It Or Not
..

THE SOLUTION

Let X be the thousand's digit,
Y be the hundreds's digit,
Z be the ten's digit,
and Q be the one's digit.

Then the final answer will appear as XYZQ.

Eq. 1.  X + Z  =  Y + Q

Eq. 2.    X Q  =  10 Y Z

Eq. 3.  X + Y  =  ½ ( Z + Q )

Since letters are in all the plus and minus
parts of the three equations, 0000 becomes
a solution.

Arbitrarily isolating the X in Eq. 1 yields

Eq. 1b.  X  =  Y + Q - Z

Substituting this into Eq. 3 yields

Y + Q - Z + Y  =  ½ ( Z + Q )

which simplifies to

2Y + ½ Q  =  3/2 Z

Multiplying by 2 yields

4 Y + Q  =  3 Z

Isolating the Q yields

Eq. 4.  Q  =  3 Z - 4 Y

¤»¥«¤»§«¤»¥«¤»§«¤»¥«¤»§«¤

Now substitute Eq. 1b into Eq. 2 yields

( Y + Q - Z ) Q  =  10 Y Z

Now substitute Eq. 4 into this yields

( Y + (3 Z - 4 Y) - Z ) (3 Z - 4 Y)  =  10 Y Z

which simplifies to

( 2 Z - 3 Y ) ( 3 Z - 4 Y )  =  10 Y Z

Multiplying the brackets

6 Z² - 8 YZ - 9 YZ + 12 Y²  =  10 YZ

which simplifies to

6 Z² - 27 YZ + 12 Y²  =  0

Dividing by 3 yields

2 Z² - 9 YZ + 4 Y²  =  0

Factoring produces

( 2Z - 1Y ) ( Z - 4Y ) = 0

Either 2z - 1Y = 0  or  Z - 4Y = 0

¤»¥«¤»§«¤»¥«¤»§«¤»¥«¤»§«¤

Trying 2z - 1Y = 0 then Y = 2Z

If Z = 1, or 2, or 3, or 4, then Y = 2, or 4, or 6, or 8 respectively.

Taking the pair Z = 1 and Y = 2 we substitute into Eq. 4

Q  =  3(1) - 4(2) yields a negative integer, which is impossible.

All the other pairs similarly product negative Qs,

so 2z - 1Y = 0 produces no real answers.

¤»¥«¤»§«¤»¥«¤»§«¤»¥«¤»§«¤

Trying Z - 4Y = 0 then Z = 4 Y

If Y = 1, or 2, then Z = 4, or 8 respecdtively.

Taking the pair Y = 1 and Z = 4 we substitute into Eq. 4

Q  =  3(4) - 4(1) yields = 8

Substituting into Eq. 1b

X  =  1 + 8 - 4 = 5

Then the final answer, XYZQ, will be 5148.

¤»¥«¤»§«¤»¥«¤»§«¤»¥«¤»§«¤

Taking the pair Y = 2 and Z = 8 we substitute into Eq. 4

Q  =  3(8) - 4(2) yields = 16 which is too large.

Therefore the answer is 5148.

.
.