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2 years ago

how many 4 digit numbers n have the property that the 3 digit number obtained by removing the leftmost digit is one ninth of n

Mathematics

1 answer:

Tom [10]2 years ago

8 0

Answer:

Step-by-step explanation:

We want to find the number 4-digit of positive integers n such that removing the thousands digit divides the number by 9.

Let the thousands digit be 'd'. Then we want to find the integer solutions to ...

n -1000d = n/9

n -n/9 = 1000d . . . . . . add 1000d -n/9

8n = 9000d . . . . . . . . multiply by 9

n = 1125d . . . . . . . . . divide by 8

The values of d that will give a suitable 4-digit value of n are 1 through 7.

When d=8, n is 9000. Removing the 9 gives 0, not 1000.

When d=9, n is 10125, not a 4-digit number.

There are 7 4-digit numbers such that removing the thousands digit gives 1/9 of the number.

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2 years ago

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Harman [31]

I think you meant to say

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Since both the numerator and denominator are continuous at t = 2, the limit of the ratio is equal to a ratio of limits. In other words, the limit operator distributes over the quotient:

$\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)}$

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$\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)} = \frac{2^4-6}{2\cdot2^2-3\cdot2+7} = \boxed{\frac{10}9}$

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