BTGmoderatorDC wrote: ↑Tue Feb 04, 2020 7:06 pm
If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundreds digit of x a perfect cube?
(1) x = (5a+b)(5a-b)
(2) a^2 + 84 = 20a
OA
C
Source: Veritas Prep
The only perfect cubes in 0-9 are 0,1,8, so the question is asking whether the hundredth digit is one of them.
Lets first examine option B which seems an easy one.
B) We do not know what is X in terms of 'a' and 'b'; NS
a^2 + 84 = 20a ; A=14,6;
even if we know what X equals from option A, as we do not have information about 'b' this statement is insufficient. \(\Large{\color{red}\chi}\)
A) x = (5a+b)(5a-b) ; x=25a^2 - b^2;
a=3, b=5 --> x=200; is hundred th digit 0/1/8 ? NO
a=6, b=4---> x=884; is hundred th digit 0/1/8 ? YES
So, option A is also insufficient. \(\Large{\color{red}\chi}\)
Let's combine A&B
a=14,6 in both
x= 4900-b^2 when a=14
x= 900-b^2 when a=6
In both cases we know the hundredth digit is 8; \(\Large{\color{green}\checkmark}\)
Therefore,
C is the correct answer.
