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

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)

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

1154 = 1155 - 1 = 11k - 1, for some positive value k
& 1156 = 1155 + 1 = 11k + 1

So, 1154*1156 = (11k - 1)(11k + 1) = (11k)^2 - 1 = 11(11k^2) - 1 = 11(n + 1) - 1 = 11n + 11 - 1 = 11n + 10
--> f(n) = 1154*1156 can be expressed of the form 11n + 10 for some value of 'n'

IMO Option E

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

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

For could be true questions, we only need to prove that the function is true for just one case.

1154*1156 = 1,324,024

Testing with f(n)=3n-2
3n-2=1,334,024
3n=1,334,022
n=444,674
Since f(n) is defined for integers and the function f(n)=3n-2 yields an integer when equated to 1154*1156, then f(n) is true for 3n-2, when n=444,674

The answer is option A.

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

\(f(n)=1154*1156=1155^2-1^2=1155^2-1\)

A. \(3n−2=1155^2-1…3n=1155^2+1…1155=factor(3)…1≠factor(3)\)
B. \(5n−2=1155^2-1…5n=1155^2+1…1155=f(5)…1≠f(5)\)
C. \(5n+3=1155^2-1…1155=f(5)…-1-3=-4…-4≠f(5)\)
D. \(7n−2=1155^2-1…1155=f(7)…-1+2=1…1≠f(7)\)
E. \(11n+10=1155^2-1…1155=f(11)…-1-10=-11…-11=f(11)…f(n)=11n+10=valid\)

Ans (E)

IMO, Ans A.

Both 1154 and 1156 when divided by 3 leaves a remainder of 2.

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

\(1155^{2} —1= 11*11*105*105—1\)

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

The answer is E

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f(n)=1154∗1156=(1155−1)∗(1155+1)=1155^2−1

A. 3n−2
B. 5n−2
C. 5n+3
D. 7n−2
E. 11n+10

For all options, put ->
3n - 2 = 1155^2−1
or
5n - 2 = 1155^2−1
and so on.

Except for E - 11n+10, you'll get a fraction for every other answer and since n is an integer, E is the answer.

Please Kudos

1154/3, Remainder= 2
1156/3, Remainder= 1
So, 1154*1156/3, Remainder= 2*1= 2
Therefore 3n+2 or 3n-1 can be the functions and not 3n-2 as we need to remove this Remainder to make it divisible by 3.

Similarly,
1154/5, Remainder= 4
1156/5, Remainder= 1
So, 1154*1156/5, Remainder= 4*1= 4
Therefore 5n+4 or 5n-1 can be the functions and not 5n-2 or 5n+3 as we need to remove this Remainder to make it divisible by 5.

Similarly,
1154/7, Remainder= 6
1156/7, Remainder= 1
So, 1154*1156/7, Remainder= 6*1= 6
Therefore 7n+6 or 7n-1 can be the functions and not 7n-2 as we need to remove this Remainder to make it divisible by 7.

Now,
1154/11, Remainder= 10
1156/11, Remainder= 1
So, 1154*1156/11, Remainder= 10*1= 10
Therefore 11n+10 or 11n-1 can be the function as we need to remove this Remainder to make it divisible by 11.

Hence, Ans E

chondro48 You deserve way better than Q49

I do agree.
Some of his explanation are very good.

_________________

Each answer is written in terms of the remainder equation, in which the coefficient is the divisor

Braking the output down to its prime factorization:

(2)^3 * (17)^2 * 577

When you divide the output by 11, does it leave a remainder of 10?

(8 / 11)Rof

*

(17^2 / 11)Rof

*

(577 / 11)Rof

= excess remainder =

8 * (6)^2 * 5 =

40 * 36 ——-> divide through by 11 again to remove the excess remainder

(40 / 11)Rof = 7
*
(36 / 11)Rof = 3

= 21 excess remainder

21 is (-1) shy from the next multiple of 11 (which is 22)

Thus, the output, when divided by 11, yields a remainder of 10

(E)
11n + 10 could describe the function

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