1. ## An Easy-Hard Question

The mystery # is a decimal....

-The digit in the 1's place is less then then digit in 10th place..
-No 2 digits are the same..
-The number is less than the # of years in a century...
-The digit in the 10th place is less than the digit in the 1000's...
-There are 2 digits in the whole part..
-There are 5 digits all together...
-The sum of the digits is 20....
-No digit is odd...
-The product of the digits in the ones place and the 100th is zero...
-The sum of the digits in the ones place and the 100th is two...
-The number is greater than 5 dozen...
-The sum of the digits in the decimal part of the # is the same as the digits in the whole part of the number.
-The product of the digits in the 10ths place and the 1000's is equal to 2 dozen...
-The product if the digits in the 100's place and the tens place is zero..

-What is the mystery # ________

2. Let us begin with these pieces:
- The number is less than the # of years in a century...
- There are 2 digits in the whole part..
- There are 5 digits all together...
- The number is greater than 5 dozen...

From this given, we can establish that:

$\displaystyle x = 10a + b + 0.1c + 0.01d + 0.001e$

$\displaystyle 60<x<100$

Now, let us translate the rest of the given into algebraic terms:

$\displaystyle b<c$

$\displaystyle c<e$

$\displaystyle a+b+c+d+e=20$

$\displaystyle bd=0$

$\displaystyle b+d=2$

$\displaystyle c+d+e=a+b$

$\displaystyle ce = 24$

$\displaystyle cd=0$

1. Now, step number 1: solve simultaneously:

$\displaystyle a+b+c+d+e=20$

$\displaystyle a+b-c-d-e = 0$

to get $\displaystyle a+b = 10$

2. Now solve simultaneously again with:

$\displaystyle a+b = 10$

$\displaystyle b+d = 2$

to get $\displaystyle a-d = 8$

3. Then, let us examine the given:

$\displaystyle ce = 24$

$\displaystyle cd=0$

It is obvious to see that d = 0. It is impossible that c = 0 since the product of c and e is 24. This is the key; plug in the relations we got previously:

4.
$\displaystyle \mbox{d = 0}$

$\displaystyle \mbox{a = 8}$

$\displaystyle \mbox{b = 2}$

$\displaystyle a+b+c+d+e=20$

$\displaystyle \implies c+e = 10$

$\displaystyle \implies c+\frac{24}{c} = 10$

$\displaystyle \implies c^2 - 10c + 24 = 0 \implies (c-4)(c-6)=0$

If c = 6, then e = 4 but this doesn't satisfy the inequality $\displaystyle c < e$
Therefore, c = 4 and e = 6

$\displaystyle \therefore x = 10(8)+2+0.1(4)+0+0.001(6) = 82.406$

3. I'm getting all confused with "whole part" and "all together" and also with digit.
So I didn't bother to solve the problem. Anyway, I have a question from
-The digit in the 10th place is less than the digit in the 1000's...
and
82.406
Are there 10 places for digits? 1000? In 82.406 of course.

4. The whole part usually refers to the part of the number that is not a decimal/ left to the decimal point. Take 123.45 for instance:

$\displaystyle 123.45 = 1(100) + 2(10) + 3(1) + 4(0.1) + 5(0.01)$

123 is the whole part. 1 occupies the hundred place, 2 occupies the ten place, 3 occupies the one place, 4 occupies the tenth place, and so on. To refer to a place that is left of the decimal, we follow the same naming process as we did to digits to the left of the decimal but we add the suffix -th to indicate a fraction.

$\displaystyle \text{Tenth} \implies \frac{1}{10}$

$\displaystyle \text{Hundredth} \implies \frac{1}{100}$

$\displaystyle \text{4~Hundredths} \implies \frac{4}{100}$

For 82.406, there is a tens place. There is no 1000 place, but there is a 1000th place.

5. Thank you Chop Suey.
-The digit in the 10th place is less than the digit in the 1000's...
should 1000's be replaced by 1000th? Hmm, maybe not. I don't understand well, nevermind.

6. Originally Posted by arbolis
Thank you Chop Suey.
I'm still confuse about should 1000's be replaced by 1000th? Hmm, maybe not. I don't understand well, nevermind.
Actually, you are right. The 1000's should be replaced by 1000th. The problem may not have been copied in a consistent manner.