# How to solve these type of problems?

Dear All, I need your help..... Please tell me how to solve these types of problems LOGICALLY..!!! QUE. Can you arrange 9 numerals - 1, 2, 3, 4, 5, 6, 7, 8 and 9 - (using each numeral just once) above and below a division line, to create a fraction equaling to 1/3 (one third)? Expecting your valuable comments......!!!!!!!

Note by Akhil K
5 years, 8 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Congratulations on being my 300th follower! Great problem! I have no idea, however.

- 4 years, 8 months ago

5832/17496

- 5 years, 8 months ago

5832/17496

- 5 years, 8 months ago

thanks for valuable comment.Please let me know HOW you solved it.

- 5 years, 8 months ago

5823/17469

- 5 years, 8 months ago

Dear Yonatan, Please share me HOW you solved it?

- 5 years, 8 months ago

It's easy to see the numerator will have four digits, the denominator five. Call the digits abcd/efghi. Also, abcd is divisible by 3, since a+b+c+d+e+f+g+h+i is divisible by 3, and e+f+g+h+i is divisible by 3 (since efghi is) so a+b+c+d must be as well. . Because efghi > 12345, we know abcd >= 4115 (so also >= 4123). This immediately eliminates a=1,2, or 3.

Now I made an assumption that if I will be able to solve this problem quickly, there should be relatively few carries, If there are no carries (besides the last digit), {b,c,d} = {1,2,3}. It's easy to check that there are no solutions of this form, so we must make a weaker assumption, that there is just 1 carry. The options for {c,d} are {{1,2}, {2,3}, {1,3}}. The first pair is eliminated because e is either 1 or 2, so let's try {2,3}. ab23 = efg69. We know a+b = 1 (mod 3) and a >= 4. Now there aren't many pairs left to try, so just guess and check until we find the smallest solution of the presumed form, which is 5823.

- 5 years, 8 months ago

congrats dear Yonatan....!!!!! This type of procedure,i had been waiting.Thanks a lot!!!!!!!!!!!!!!!!!!!!!

- 5 years, 8 months ago