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Mathematics-Online problems: | ||

## Solution to the problem of the (previous) week |

** Problem:**

*Hint:* Different letters represent different digits.

**Answer:**

** Solution:**

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

10 = T + U > 2A + A = 3A
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
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
for B
we have **B=8**.

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

[problem of the week]