math kangaroo need help please and thank you

[T W O]

Two 3-digit numbers have all their 6 digit distinct. The first digit of the second number is twice the last digit of the first number. What is the smallest possible sum of the two such numbers?

 

[lev888]

Is this a practice problem?

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[T W O]

Is this a practice problem?

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Welcome to FreeMathHelp.com! Please take the time to read the following before you make your first post. It will help you to get your math questions answered promptly and in the most helpful manner. Don't post a list of homework problems. Please start new threads for new exercises. Try to...

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yes

 

[Dr.Peterson]

Two 3-digit numbers have all their 6 digits distinct. The first digit of the second number is twice the last digit of the first number. What is the smallest possible sum of two such numbers?
View attachment 24831

All you can do here is to think about how you make a three-digit number as small as possible, and try different possibilities. Don't forget that one of the numbers may contain a zero.

This is the sort of problem for which it's hard to give hints more specific than that without seeing your work.

 

[T W O]

All you can do here is to think about how you make a three-digit number as small as possible and try different possibilities. Don't forget that one of the numbers may contain a zero.

This is the sort of problem for which it's hard to give hints more specific than that without seeing your work.

Thank you so much for your help

 

[lev888]

Thank you so much for your help

Start with the 2 digits mentioned in the question. What's the smallest combination? What's the next smallest? Does the second allow you to make the sum smaller? How about the third possibility? Is it obvious that it's too big?

 

[Steven G]

Show us you work so that we can help you.

Let the first of the three digit number be 100a + 10b + c.
The 2nd three digit number must be 100(2c) + 10d + e
a, b, c, d and e must all be different and a & c are not 0.

You obviously want a and 2c to be as small as possible.