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)
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.)
The \(KUB\) is a cube. Prove that the ball, \(CIR\), is not a cube. (\(KUB\) and \(CIR\) are three-digit numbers, where different letters denote different numbers).
Can I replace the letters with numbers in the puzzle \(RE \times CTS + 1 = CE \times MS\) so that the correct equality is obtained (different letters need to be replaced by different numbers, and the same letters must correspond to the same digits)?
In the entry \({*} + {*} + {*} + {*} + {*} + {*} + {*} + {*} = {*}{*}\) replace the asterisks with different digits so that the equality is correct.
It is known that \(AA + A = XYZ\). What is the last digit of the product: \(B \times C \times D \times D \times C \times E \times F \times G\) (where different letters denote different digits, identical letters denote identical digits)?