Question

1) a+b+c, если abb+bcc+cca=1332​

Answer 1

Ответ:

Давайте сначала проанализируем суммы, которые вы дали:

100a+10b+b+100b+10c+c+100c+10a+a=1332

111a+111b+111c=1332

111(a+b+c)=1332

разделите каждую сторону на 111, получится 12

Объяснение:

вот

Answer 2

To find the value of $a + b + c$ given the equation $abb + bcc + cca = 1332$, we need to solve for the individual values of $a$, $b$, and $c$.

The given equation is: $abb + bcc + cca = 1332$

Let's analyze this equation:

• $abb$ suggests $a$ repeated twice followed by $b$.
• $bcc$ suggests $b$ repeated twice followed by $c$.
• $cca$ suggests $c$ repeated twice followed by $a$.

Since these are concatenations of the digits, we can rewrite the equation as: $100a + 10b + b + 100b + 10c + c + 100c + 10a + a = 1332$

Simplifying this equation, we get: $122a + 111b + 111c = 1332$

Now, we can rearrange this equation to solve for $a + b + c$: $a + b + c = \frac{1332}{122 + 111 + 111}$

Calculating this gives: $a + b + c \approx \frac{1332}{344} \approx 3.872$

However, since $a$, $b$, and $c$ are digits, we'll round this to the nearest integer, which is $4$.

So, $a + b + c = 4$.