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Mathematics-Online problems:

Solution to the problem of the (previous) week

Problem:

What are the digits represented by the letters in the following pattern of a ,,written`` multiplication of two three-digit numbers?

{\large \begin{tabular}{ccc ccc ccc ccc c} A&B&I&*&T&U&R \\ [+0.5ex] \hline ... ...&I&U&E \\ [+0.5ex] &C&A&H&E \\ [+0.5ex] \hline &C&U&S&E&S&I \end{tabular} }

Hint: Different letters represent different digits.


Answer:

$ S=$          $ U=$          $ C=$          $ H=$                 
                       
$ A=$          $ R=$          $ B=$          $ E=$          $ I=$          $ T=$         

Solution:

The first digit to be identified is $ 0.$ In the last but one column, the sum S+E=S does not have a carry over from the last column. Therefore, E=0.

So, the last digits of the products T$ \ast$I and U$ \ast$I are both $ 0.$ Since none of the factors concerned is $ 0,$ we obtain 0 as a result only if $ 5$ is multiplied by an even number. Both products have the common factor I, which has to be $ {\bf 5}$ (and the other two factors T and U have to be even).

Now, the pattern looks as follows:

{\large \begin{tabular}{ccc ccc ccc ccc c} A&B&5&$\ast$&T&U&R \\ [+0.5ex] \... ...&U&0 \\ [+0.5ex] &C&A&H&0 \\ [+0.5ex] \hline &C&U&S&0&S&5 \end{tabular} }

For the determination of further digits, we consider the auxiliary products. We note that all three four-digit products CTSI, HIUE, and CAHE result from a multiplication of the three-digit number ABI with a one-digit number and thus have to be less than ABI$ \ast 10$. Therefore, the thousands digits of the auxiliary products cannot be greater than A, and so we have C < A, H < A, and particularly A > 2.

Since two of the auxiliary products have the same thousands digit, their hundreds digits must differ at least by A. Hence, we must have either T$ \leq$A - A = 0 or T$ \geq$A + A = 2A. The first case can be ruled out, since E=0.

The fifth column from the right does not produce a carry over. So, the sum U is greater than each of the two summands, particularly U > A.

If we look at the third column from the right, we find that the addition generates a carry over. Consequently, 10 = T + U, leading to

10 = T + U > 2A + A = 3A $ \Rightarrow$ A < 4
taking the inequalities from above into account.

Thus, A > 2 and A < 4, which means A=3. Due to the constraints U > A = 3 and T$ \geq$2A = 6, the equation 10 = T + U allows nothing else but U=4 and T=6. So, the pattern now looks like

{\large \begin{tabular}{ccc ccc ccc ccc c} 3&B&5&$\ast$&6&4&R \\ [+0.5ex] \... ...&4&0 \\ [+0.5ex] &C&3&H&0 \\ [+0.5ex] \hline &C&4&S&0&S&5 \end{tabular} }

As mentioned above, both H and C have to be less than A, and in the fifth column from the right we now get H=1, and thus C=2.

Including the carry over from the third column from the right, the fourth column from the right produces S = C + I + H + 1 = 2 + 5 + 1 + 1, so S=9. Finally, there is only one odd digit remaining for R (the last digit of a product is 5 if one of the factors ends by 5), so R=7, and for B we have B=8.

(These results are confirmed by the divisions CAHE/T = 2310/6 = 385 and CUSESI/ABI = 249095/385 = 647.)

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