Long Multiplication (Puzzle)

[BigBeachBanana]

[math]\begin{array}{r}AB\\ \times BA \\ \hline 3\,\,8\,\,4\\ 2\,\,5\,\,6\,\,0\\ \hline 2\,\,9\,\,4\,\,4\\ \end{array}[/math]Given [imath]A[/imath] and [imath]B[/imath] are different values. Find [imath]A[/imath] and [imath]B[/imath].

 

[Otis]

Spoiler

A must be the larger digit

A times B must end in 4

Product 14 too small to generate 384, but 24 works

A = 6
B = 4

?

[imath]\;[/imath]

 

[Steven G]

Spoiler

A=1,B=4
A=2, B=7
A=3, B=8
A=4, B =6
The above are the only choices to consider.
A=4 and B=6 work

 

[Otis]

? Hey Steve, note the rows 384 and 2560 in the op.

[imath]\;[/imath]

 

[Dr.Peterson]

? Hey Steve, note the rows 384 and 2560 in the op.

[imath]\;[/imath]

I wonder if you learned a different way to do multiplication than I did. My way, the second partial product would be B times AB, implying that B > A.

Of course, it makes no difference in what the product is.

 

[lookagain]

I wonder if you learned a different way to do multiplication than I did. My way, the second partial product would be B times AB, implying that B > A.

Of course, it makes no difference in what the product is.

AB*A = 384 and AB*B = 256, as opposed to 2560.
Then A > B.

 

[Dr.Peterson]

AB*A = 384 and AB*B = 256, as opposed to 2560.
Then A > B.

Oops. That's why I don't write the 0 ...

 

[lex]

Spoiler

[math]\dfrac{A}{B}=\dfrac{384}{256}=\dfrac{3}{2} \Rightarrow \text{ 3, 2 or 6, 4 }[/math]
or more offbeat:
[math]AB \times (A - B)=384-256=128=2^7\\ \text{+ as } \therefore AB \text{ is a power of 2, as is } (A-B): 16, 32, 64 \Rightarrow 64[/math]

 

[Dr.Peterson]

Actually, my first thought when I saw it was just to factor the product:

Spoiler

[imath]2944 = 2^7\times 23[/imath]
To get two two-digit factors, they have to be [imath]23\times 2=46[/imath] and [imath]2^6=64[/imath].
Since the digits work, we're done.

Or, to make sure the partial products work, factor them and look for a common factor:

Spoiler

[imath]384=2^7\times3[/imath]
[imath]256=2^8[/imath]
We need a two-digit common factor multiplied by two single digits, so one number is [imath]2^6=64[/imath] and the others are [imath]6[/imath] and [imath]4[/imath].
[imath]2^7=128[/imath] is too big, and [imath]2^5=32[/imath] is too small (leaving the "digits" as [imath]2^2\times3=12[/imath] and [imath]2^3=8[/imath]).

But those seemed too different from the usual style of these puzzles.

 

[lex]

@Dr.Peterson
Nice analysis!

 

[Otis]

I interpret the puzzle's goal to be: Find A and B to match the clue as posted.

Spoiler

(Readers may place a zero in the units column, as they each see fit.)

 

[JeffM]

I went

[math]25 < A^2 < 49 \implies A = 6. \\ \therefore 6(60 + B) = 384 \implies 6B = 24 \implies B = 4.\\ \therefore AB = 64.\\ BA = 46.\\ 6 * 64 = 384.\\ 40 * 64 = 2560. \ \checkmark[/math]

 

[BigBeachBanana]

Just some background. This problem is typically presented to 9th graders as a challenge. It's interesting to see different approaches to the problem at different levels. All roads lead to Rome.?

 

[Otis]

This problem is typically presented to 9th graders as a challenge.

All the more reason to be precise!

?

[imath]\;[/imath]

 

[Steven G]

? Hey Steve, note the rows 384 and 2560 in the op.

[imath]\;[/imath]

I see now that I took the long way.