MHB Finding an unknown 3-digit number

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Multiplying a 3-digit number by 1002 results in the format AB007C, where A, B, and C are digits. The discussion reveals that using the distributive property and testing various 3-digit numbers has not yielded a solution. Analysis shows that for the last digit C to be valid, it must equal 0, leading to a contradiction in the fifth column where 2B must end in 7, which is impossible since 2B is even. Therefore, the conclusion is that this alphametic problem has no solution. The original number cannot be determined under the given conditions.
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If you multiply a 3-digit number by 1002 and get AB007C, where A, B, and C stand for digits, what is teh original 3-digit number? We've tried using the distributive property as well as just multiplying by 3-digit numbers that give you a 0 in the thousandths place but no luck. Thanks for any suggestions!
 
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Hello, mklilly2000!

This alphametic has no solution.

If you multiply a 3-digit number ABC by 1002 and get AB007C,
what is the original 3-digit number?
Arrange the multiplication like this:

\begin{array}{ccccc}<br /> ^1&amp;^2&amp;^3&amp;^4&amp;^5&amp;^6 \\<br /> &amp;&amp;&amp; A&amp;B&amp;C \\<br /> \times &amp;&amp; 1 &amp; 0&amp; 0 &amp; 2 \\ \hline<br /> &amp;&amp;&amp; 2A &amp; 2B&amp; 2C \\<br /> A&amp;B&amp;C \\ \hline<br /> A&amp;B&amp;0&amp;0&amp;7&amp;C \end{array}

In column-6, we see that 2C ends in C.
Hence, C=0 and there is no "carry" to column-5.

In column-5, we see that 2B ends in 7.
Since 2B is even, this is clearly impossible.

Q.E.D.

 
Soroban, the problem does not say that the original number is ABC.
 
Hint:

The middle digit of the three-digit number is a 3.
 
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