Find the least number such that
8x7y5
is divisible by 75.
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Because it had too many problems to solve, just like your question! But don't worry, I'm here to help you out.
To find the least number such that 8x7y5 is divisible by 75, we need to determine the value of x and y.
To be divisible by 75, a number must be divisible by both 5 and 15 (which is the product of 3 and 5).
Divisibility by 5:
For a number to be divisible by 5, the units digit must be either 0 or 5. Looking at the given number, the units digit is 5. Therefore, y = 5.
Divisibility by 15:
For a number to be divisible by 15, it must be divisible by both 3 and 5.
To check divisibility by 3, we can add up the digits of the number. In this case, 8 + 7 + 5 = 20. Since 20 is not divisible by 3, we need to increase the value of x to make it divisible by 3.
The next multiple of 3 that is greater than 20 is 21. So we need to choose the value of x that makes 8 + 7 + 5 + x + 1 divisible by 3. The value of x that satisfies this condition is x = 1.
Therefore, the least number that is divisible by 75 is 8715.
To find the least number such that
8x7y5 is divisible by 75, we need to determine the values of x, y, and any other missing digits.
We know that a number is divisible by 75 if it is divisible by both 5 and 15 (since 5 * 15 = 75).
For a number to be divisible by 5, the last digit must be either 0 or 5.
Therefore, we have two possibilities for y: 0 or 5.
Next, let's consider divisibility by 15. A number is divisible by 15 if it is divisible by both 3 and 5.
To determine if 8x7y50 is divisible by 3, we can check the sum of its digits. The sum of the digits is given by:
8 + x + 7 + y + 5 + 0 = 20 + (x + y).
For 8x7y50 to be divisible by 3, the sum (20 + x + y) must also be divisible by 3.
If y = 0, then the sum becomes (20 + x), and for divisibility by 3, (20 + x) must be divisible by 3.
If y = 5, then the sum becomes (20 + x + 5), and for divisibility by 3, (25 + x) must be divisible by 3.
Now, let's calculate the possible values of x.
If y = 0, then (20 + x) can be divisible by 3 if x is 1 or 4 (since 21 and 24 are divisible by 3).
If y = 5, then (25 + x) can be divisible by 3 if x is 1 or 7 (since 26 and 32 are divisible by 3).
So, the possible values of x are 1, 4, and 7.
Now we can determine the least number by considering the combinations of x and y:
1. If y = 0 and x = 1, the number is 81750.
2. If y = 0 and x = 4, the number is 84750.
3. If y = 0 and x = 7, the number is 87750.
4. If y = 5 and x = 1, the number is 81550.
5. If y = 5 and x = 4, the number is 84550.
6. If y = 5 and x = 7, the number is 87550.
Out of these possible combinations, we need to find the least number. The smallest number is formed when x = 1 and y = 0. Therefore, the least number such that 8x7y5 is divisible by 75 is 81750.