If you multiply me by 2, I become a number greater than 20 and less than 40. If you multiply me by 6, I end in 8
Multiply me by 4, I end in 2
are the last two numbes 4 and 2? or 40 and 20?
a b c
nevermind my other comment, the answe is 13.
13*2=26
13*6=78
the answer is 13.
13*2=26 which is greater than 20 but less than 40.
13*6=78 which ends in 8.
13*4=52 which ends in 2.
Hope this Helped!
it could be 23 28 33 or 38
or the answer is 18.
18*2=36 which is greater than 20 but less than 40.
18*6=108 which ends in 8.
18*4=72 which ends in 2.
To solve this problem, let's start by assuming the number is represented by "x".
According to the first statement, if you multiply "x" by 2, the result should be greater than 20 and less than 40. So we can write the following inequality:
20 < 2x < 40
Next, let's consider the second statement. If you multiply "x" by 6, the result ends in 8. This means the last digit of the result must be 8. We can express this mathematically as:
6x ≡ 8 (mod 10)
Lastly, according to the third statement, if you multiply "x" by 4, the result ends in 2. This means the last digit of the result must be 2. We can express this mathematically as:
4x ≡ 2 (mod 10)
Now, let's solve these equations step by step:
For the first statement, divide all parts of the inequality by 2:
10 < x < 20
This tells us that "x" must be between 10 and 20.
For the second statement, we can simplify the equation by dividing both sides by 2:
3x ≡ 4 (mod 5)
To find a solution to this congruence, we can check each possible residue of "x" modulo 5 and see if it satisfies the equation (x ≡ 0, 1, 2, 3, or 4 mod 5). We find that if x ≡ 2 (mod 5), then the congruence is satisfied.
For the third statement, we can divide both sides of the equation by 2:
2x ≡ 1 (mod 5)
Again, we check each possible residue of "x" modulo 5 and find that if x ≡ 3 (mod 5), then the congruence is satisfied.
Combining all the constraints, we find that:
x ≡ 2 (mod 5)
x ≡ 3 (mod 5)
To find the solution that satisfies both congruences, we can use the Chinese Remainder Theorem (CRT). It states that if we have a system of congruences where the moduli are pairwise coprime (which is true in this case), then there exists a unique solution modulo the product of the moduli.
Using CRT, we can find the solution:
x ≡ 23 (mod 10)
Therefore, the number you are looking for is 23.