Problems

Solve the rebus \(AC \times CC \times K = 2002\) (different letters correspond to different integers and vice versa).

Can the equality \(K \times O \times T\) = \(U \times W \times E \times H \times S \times L\) be true if instead of the letters in it we substitute integers from 1 to 9 (different letters correspond to different numbers)?

Rebus. Solve the numerical rebus \(AAAA-BBB + SS-K = 1234\) (different letters correspond to different numbers, but the same letters each time correspond to the same numbers)

When Harvey was asked to come up with a problem for the mathematical Olympiad in Sunny City, he wrote a rebus (see the drawing). Can it be solved? (Different letters must match different numbers).

In the rebus below, replace the letters with numbers such that the same numbers are represented with the same letter. The asterisks can be replaced with any numbers such that the equations hold.

An explanation of the notation used: the unknown numbers in the third and fourth rows are the results of multiplying 1995 by each digit of the number in the second row, respectively. These third and fourth rows are added together to get the total result of the multiplication \(1995 \times ***\), which is the number in the fifth row. This is an example of a “long multiplication table”.

Alex laid out an example of an addition of numbers from cards with numbers on them and then swapped two cards. As you can see, the equality has been violated. Which cards did Alex rearrange?

In the line of numbers and signs \({}* 1 * 2 * 4 * 8 * 16 * 32 * 64 = 27\) position the signs “\(+\)” or “\(-\)” instead of the signs “\(*\)”, so that the equality becomes true.

The code of lock is a two-digit number. Ben forgot the code, but he remembers that the sum of the digits of this number, combined with their product, is equal to the number itself. Write all possible code options so that Ben could open the lock quickly.

Find the smallest four-digit number \(CEEM\) for which there exists a solution to the rebus \(MN + PORG = CEEM\). (The same letters correspond to the same numbers, different – different.)

Arrange brackets and arithmetic signs around these numbers so that the correct equality is obtained: \[\frac{1}{2}\quad \frac{1}{6}\quad \frac{1}{6009} \ = \ 2003.\]