|[home] [lexicon] [problems] [tests] [courses] [auxiliaries] [notes] [staff]|
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?
Hint: Different letters represent different digits.
Solution:The first digit to be identified is 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 TI and UI are both Since none of the factors concerned is we obtain 0 as a result only if is multiplied by an even number. Both products have the common factor I, which has to be (and the other two factors T and U have to be even).
Now, the pattern looks as follows:
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. 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 TA - A = 0 or TA + 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
Thus, A > 2 and A < 4, which means A=3. Due to the constraints U > A = 3 and T2A = 6, the equation 10 = T + U allows nothing else but U=4 and T=6. So, the pattern now looks like
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
we have B=8.
(These results are confirmed by the divisions CAHE/T = 2310/6 = 385 and CUSESI/ABI = 249095/385 = 647.)