Question

Задумано простое трёхзначное число,все цифры которого различны.На какую цифру оно может оканчиваться,если его последняя цифра равна сумме первых двух?

Answer 1

Нуу не четная, не 1, не 5, не 3, т.к. сумма будет 6 и не 9, то выходит 7

Answer 2

Solving the Problem

To find the possible last digit of a three-digit number where all digits are different and the last digit is equal to the sum of the first two digits, we can approach the problem systematically.

Let's denote the three digits as A, B, and C, where A is the hundreds digit, B is the tens digit, and C is the ones digit. We are given that A, B, and C are all different, and C = A + B.

Possible Values for A, B, and C

To find the possible values for A, B, and C, we can start by considering the range of values for A and B. Since A, B, and C are all different digits, the possible values for A and B are limited.

1. A can be 1, 2, or 3. 2. B can be 0, 1, 2, 3, 4, 5, 6, 7, or 8.

Calculating the Possible Values for C

Given the constraints, we can calculate the possible values for C by adding A and B for each combination of A and B.

1. If A = 1, then C can be 2, 3, or 4. 2. If A = 2, then C can be 3, 4, 5, or 6. 3. If A = 3, then C can be 4, 5, 6, or 7.

Conclusion

Based on the calculations, the possible last digit for the three-digit number, where all digits are different and the last digit is equal to the sum of the first two, can be 2, 3, 4, 5, 6, or 7.

Therefore, the possible last digit for the given conditions is 2, 3, 4, 5, 6, or 7.