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Xavier writes down a three-digit positive integer. Youssef writes a  in front of the digits of Xavier's number, so that the resulting number has four digits. If this four-digit number is  times as large as Xavier's original number, what number did Xavier write down?

 

So here's my approach:
 

we know xaviers number is in the range of 100-999

and we know it has to be a multiple of 13

and the number we are trying to find is between 1000-9999

and it has to start with a 3.

 Mar 19, 2023
 #1
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 "If this four-digit number is  times as large as Xavier's original number, what number did Xavier write down"?  How many times larger?                             

 Mar 19, 2023
 #2
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Ah sorry I meant 13 times larger. for some reason it didn't show

 Mar 19, 2023
 #6
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Try proofreading your questions Jason!

Melody  Mar 20, 2023
 #3
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Let Xavier's three-digit number be represented as $ABC$, where $A$, $B$, and $C$ are digits representing the hundreds, tens, and ones digits, respectively. Then, Youssef's four-digit number is $3ABC$.

Since the four-digit number is 13 times as large as the original number, we have:

$$3ABC = 13 \cdot ABC$$

Simplifying this equation, we get:

$$3000 + 100A + 10B + C = 13(100A + 10B + C)$$

Expanding the right side, we get:

$$3000 + 100A + 10B + C = 1300A + 130B + 13C$$

Simplifying this equation further, we get:

$$1170A + 120B = 2999$$

Since $A$ and $B$ are digits, the left-hand side of this equation is an integer between 0 and 999. Therefore, the only possible value for $A$ is 2, and the only possible value for $B$ is 4. Substituting these values into the equation, we get:

$$1170(2) + 120(4) = 2520 + 480 = 2999$$

Therefore, Xavier's original number is $ABC = \boxed{240}$.

 Mar 19, 2023
 #4
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Try 500. 6,500 is 13 times larger than 500

 

6,500 / 500 ==13 - which is what you want.

 Mar 19, 2023
 #5
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It makes sense. but the system said its wrong.
I also tried 231 which worked?

 Mar 19, 2023

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