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# Digit question

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The number A8 is a two-digit number with tens digit A and units (ones) digit 8.  Similarly, 3B is a two-digit number with tens digit 3 and units digit B. When A8 is multiplied by 3B, the result is the four-digit number C430.

If A, B, and C are each different digits from 0 to 9, determine the values of A. B, and C.

Oct 15, 2021

#1
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If both two-digit numbers A8 and 3B have two digits and the product of these factors should be a 4-digit number, then there could not be a 1 at A, because the largest possible result from the factors A8 and 3B would be 702. The following possibilities would not be possible through a small product that is smaller than 1000:

18 and 30 ... 18 and 39; 28 and 30 ... 28 and 35, so together there would be 15 possibilities that are not possible because of their small products that are less than 1000. Now note down an important point:

To get C430 as a product, the factor 3B must (except for a few exceptions) be on either one of these digits 0 or 5. This important point now gives rise to more impossible factors, namely: 28 and 30 ... 28 and 39, since we said that 3B should end in 0 or 5, it was already in the first lists that they would not be possible to form a product that is 1000 and the factor 3B should end in 0 or 5. Now, according to the new rules, there can be no 1 and no 2 in the factor A. At this moment, they are trying to deal with the factor A8, so this factor is now called 38, and as I said, the second factor 3B can only have a 0 or 5, because these factors form a product that is at least divisible by 5. 38 and 30 add up to 1140 - doesn't fit and 38 and 35 - doesn't fit either. And this is how we always proceed. Of course, I'm not writing the new rules and the steps to find possible solutions, but I'm saying that such numbers are a possibility.
These are the following possibilities:

98 and 35

I think that's the onliest possibility, where it applies:

A=9

B=5

C=3

Straight

Oct 15, 2021
#2
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Thanks Straight,

It is easiest to do these by long multiplcation:

See if you can work through what I have done.

Oct 16, 2021
#3
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Hello melody,

couldn't B be 0 because 8*0 is 0 and the solution must end with 0?

Yes, I do know that the solution is B=5, but why not?

Straight

Oct 16, 2021
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Thanks for the question straight, To be honest I had not thought about it. But:

B can't be zero because 3*8 = 24  but we would need a units digit of 3

So if B=0 then the tens digit of the answer would have to be 4.  But it is not, so b is not equal to 0

Melody  Oct 17, 2021
edited by Melody  Oct 17, 2021
#5
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Hello Melody,

sorry, but what? B can be 0

because A8 * 0 is 0 it ends with 0 and that's what we're needing.

For example A could be 5, then 58 * 30 is 1740 and this number ends with a zero,

no matter what digit A is the result ALWAYS ends with 0.

Straight

Oct 17, 2021
edited by Straight  Oct 17, 2021
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You have not understood my logic straight my logic Straight.

Try writing it out, as a multiplication, and see what happens if B=0  you will find that the answer will not be C430  it will have to be C440.

becasue 3*8=24

So B is not 0

Melody  Oct 18, 2021