If the seven-digit number 52A6B7C is divisible by 33 and A, B, C are primes then the maximum value of 3A+2B+5C?
Given number- 52A6B7C. Difference between even and odd places numbers sum will be divisible by 11 ,0 or multiple of 11 5+A+B+C – 2+6+7= 11a A+B+C – 10 =11a maximum or 0 is minimum So 3A+2B+5C = 3×3+2×5+5×2=9+10+10=29
I. 2x² - 15x + 13 = 0
II. 3y² - 6y + 3 = 0
I. 99x² + 161 x + 26 = 0
II. 26 y² + 161 y + 99 = 0
I. 35x² + 83x + 36 = 0
II. 42y² + 53y + 15 = 0
I. 18p²- 21p + 6 = 0
II. 16q² - 24q +9 = 0
I. √(74x-250 )– x=15
II. √(3y²-37y+18)+ 2y=18
I.70x² - 143x + 72 = 0
II. 80 y² - 142y + 63 = 0
I. 2x2 – 19x + 45 = 0
II. y2 – 14y + 48 = 0
I. 3x2 = 2x2 + 9x – 20
II. 3y2 = 75
A and B are the roots of equation x2 - 13x + k = 0. If A - B = 5, what is the value of k?
I. 3x2 + 3x - 60 = 0
II. 2y2 - 7y + 5 = 0
