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3 years ago

1E. The number 2A5A0A2 is divisible by 99. What is the digit A?

Mathematics

1 answer:

PilotLPTM [1.2K]3 years ago

3 0

We know, a number is divisible by 9 if the sum of digits is also divisible by 9.

Also, since A is an digit, so A is in between 0 to 9.

2+A+5+A+0+A+2 = 3A + 9

3A + 9 should be divisible by 9.

3A + 9 = 9n

A = (9n -9)/3

For n= 1 :

A = 0

For n=2 :

A = 3

For n= 3 :

A = 9

After this A > 9 which is not accepted.

Therefore, the digit A can be 0, 3 and 9 .

Hence, this is the required solution.

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following are the solution to this question:

Step-by-step explanation:

Please find the complete question in the attached file.

$\lim_{x\to 2}+f(x) = \lim_{x\to 2}+ (ax^2-bx+3) = 4a-2b+3\\\\ \to 4a-2b+3 = 4\\\\ \therefore \ \ 4a-2b = 1\\\\$

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$\to -8a + 4b = -2\\\to 10a - 4b = 3\\ \to 2a = 1\\\\ \to a = \frac{1}{2}\\\\ \to \ 4(\frac{1}{2}) - 2b = 1\\\\ \to 2 - 2b = 1\\\\ \to -2b = -1\\\\ \to b = \frac{1}{2}$

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