@shashank_prabhulet 2 numbers are 'ab' and 'cd', such that ab>cd.abcd-(ab-cd)=5481100 * ab+cd-ab+cd=548199 * ab+2 * cd=5481
99*ab and 5481 are multiples of 9, hence cd must be a multiple of 9.
At cd=18, we get ab=55
Sum of the two 2-digit numbers= 55+18=73
@hemant_malhotra said in Question Bank - Algebra - Hemant Malhotra:
Let a < b < c be the three real roots of the equation √2014 x^3 - 4029 x^2 + 2 = 0. Find b(a + c).
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_9
f(7)^2 - f(6)^2 = 517
(f(7) + f(6)) (f(7) - f(6)) = 517 = 11 x 47
As 11 and 47 cannot be further factorized and f(7) or f(6) should be natural numbers, we can say
f(7) + f(6) = 47 (Largest prime factor should come from their sum)
f(7) - f(6) = 11 (smallest prime factor should come from their difference)
solving, f(7) = 29 and f(6) = 18
f(10) = f(8) + f(9)
= f(7) + f(6) + f(7) + f(8)
= f(7) + f(6) + f(7) + f(7) + f(6)
3f(7) + 2f(6) = 3 x 29 + 2 x 18 = 123.