the last two digits = 24
options a,b,c,d wont be possible
IMO E ; 11n+10


If f(n)=1154∗1156f(n)=1154∗1156 for some integer, n, which of the following expressions could be equal to f(n)f(n)?


A. 3n−2

B. 5n−2

C. 5n+3

D. 7n−2

E. 11n+10

f(n)=1154∗1156 = 1155^2 -1

From options its visible that
1155 is divisible by 11....
so E: 1155^2 -1-10 = 1155^2 -11= which is divisible by 11

OA:E

1154x1156= 1334024

Of the five answer choices, We notice that the last term is -2 in three of the options. But after adding -2, the resulting number 1334026 is not divisible by 5 or 3 or 7. Therefore eliminate A,B and D.

We add -10 to the number and notice that the result,1334014 is divisble by 11. (divisibility test : difference between sum of odd digits and even digits should be divisble by 11)

Therefore E is the answer

\(f(n)=1154∗1156=(1155-1)*(1155+1)=1155^2-1=(3^2*5^2*7^2*11^2)-1\)

We understand that the units digit of \(f(n)=1155^2-1\) must be 4.
(B) f(n)=5n−2 --> units digit of the result is either 3 or 8
(C) f(n)=5n+3 --> units digit of the result is either 3 or 8
Thus, we confidently eliminate choices (B) and (C), since both choices NEVER yield results with units digit of 4

A. f(n)=3n−2
\(f(n)=3n−2=(3^2*5^2*7^2*11^2)-1\)
\(3n=(3^2*5^2*7^2*11^2)+1\)
\(n=(3*5^2*7^2*11^2)+1/3\)
--> \(n\) is NOT an integer, so we confidently eliminate choice (A), since \(n\) has to be some integer

D. f(n)=7n−2
\(f(n)=7n−2=(3^2*5^2*7^2*11^2)-1\)
\(7n=(3^2*5^2*7^2*11^2)+1\)
\(n=(3^2*5^2*7*11^2)+1/7\)
--> \(n\) is NOT an integer, so we confidently eliminate choice (D), since \(n\) has to be some integer

E. 11n+10
\(f(n)=11n+10=(3^2*5^2*7^2*11^2)-1\)
\(11n=(3^2*5^2*7^2*11^2)-11\)
\(n=(3^2*5^2*7^2*11)-1\)
--> \(n\) is an integer, so we are confident that choice (E) is the CORRECT ANSWER


Final answer is (E)

\(f(n) = 1154*1156= (1155 —1)(1155 +1)= 1155^{2}—1\)

\(1155^{2} —1= 11*11*105*105—1\)

E) \(11n + 10= 11n + 11—1=
11(n+1)—1\)

The answer is E

Posted from my mobile device