whats the answer to if you multiply me by 2 im greater than 20 but less than 40, if you multiply me by 4 i end in 2, and if you multiply me by 6 i end in 8
you are a 13
13 *2 = 26 (more than 20 and less than 40)
13 *4 = 52 (no restriction)
13 *6 = 78 (no restriction)
To find the answer to the given question, we need to solve a system of equations.
Let's represent the unknown number as "x".
According to the information provided:
1. "If you multiply me by 2, I'm greater than 20 but less than 40."
This can be expressed as: 20 < 2x < 40.
2. "If you multiply me by 4, I end in 2."
This implies that when 4x is divided by 10, the remainder is 2. In other words:
4x ≡ 2 (mod 10).
3. "If you multiply me by 6, I end in 8."
Similarly, this means when 6x is divided by 10, the remainder is 8. In other words:
6x ≡ 8 (mod 10).
Now, we can solve this system of equations:
1. From "20 < 2x < 40," divide all sides by 2 to get: 10 < x < 20.
2. Rewrite the congruence equation "4x ≡ 2 (mod 10)" as:
4x = 10n + 2, where n is an integer.
Divide both sides by 2: 2x = 5n + 1.
Since the right side of the equation is odd, the left side (2x) must also be odd.
Thus, x must be an odd number.
To narrow down the possible solutions, we can substitute odd numbers (10k + 1) into the equation and check if it satisfies the conditions.
Let's start with k = 0:
For k = 0, x = 10(0) + 1 = 1.
Multiply 1 by 4: 1 * 4 = 4 (does not end in 2).
Continue testing with k = 1:
For k = 1, x = 10(1) + 1 = 11.
Multiply 11 by 4: 11 * 4 = 44 (ends in 4, not 2).
We can conclude that the values of x that satisfy the condition 2 are not multiples of 10.
3. Next, let's solve the third congruence equation "6x ≡ 8 (mod 10)."
Rewrite the equation as: 6x = 10n + 8, where n is an integer.
Divide both sides by 2: 3x = 5n + 4.
Similar to the previous case, let's substitute odd numbers (10k + 4) into the equation and check for valid solutions.
For k = 0, x = 10(0) + 4 = 4.
Multiply 4 by 6: 4 * 6 = 24 (does not end in 8).
Proceeding to k = 1, x = 10(1) + 4 = 14.
Multiply 14 by 6: 14 * 6 = 84 (ends in 4, not 8).
We can conclude that the values of x that satisfy the condition 3 are not multiples of 10.
By analyzing the conditions, we find out that none of the numbers satisfy all the given conditions. Therefore, there is no specific answer to the question as stated.