x + y =36 eq.1
x = 3y eq.2
Substitute value of x (eq.2) to equation 1:
3y + y = 36
4y = 36
y=9
x = 3y
x = 3(9)
x = 27
from equation 1: x + y =36
27 + 9 = 36
Let unknown numbers be 2x and x: 2x+x = 24 3x = 24 x = 8 Therefore the two numbers are 8 and 16
33 66+33=99
12 and 12, whose squares will be 144 each. If either of the numbers is smaller than 12, then the other will be larger than 12 and its square will be larger than 144.
67
The three numbers are 2, -3, and 6.
11 & 44
1st number=12 2nd number=24
The answer is X-13 + Y- 52 = Sum 65
Let unknown numbers be 2x and x: 2x+x = 24 3x = 24 x = 8 Therefore the two numbers are 8 and 16
33 66+33=99
Let x=smaller #, 4x=larger # x+4x=45 5x=45 Divide both sides by 5. x=9 4x=36
That happens when one of the numbers is a factor of the other.
In that case, one of the numbers is a multiple of the other.
No, because the square of a number is that number times itself, so a no two numbers will have the same square.
Numbers for which the sum of the digits is divisible by 9. This is also true for 3. There are other divisibility/multiple tests for other numbers (e.g., numbers that are divisible by 5 end in 5 or 0; numbers whose last two digits are divisible by 4 are divisible by 4)
900 - 30 times 30
2 and any other even number.