For any positive integer n, let f(n) = n(n + 1) if n is even, and f(n)

Question

For any positive integer n, let  f(n) = n(n + 1)  if n is even, and  f(n) = n + 3  if n is odd. If m is a positive integer such that  8f(m + 1) - f(m) = 2 , then what is the value of m ?

A. 1
B. 3
C. 4
D. 6
E. 10

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CareerGeek · Answer

Case 1: m = even
 8f(m + 1) - f(m) = 2 
-->  8(m + 4) - m(m + 1) = 2 
-->  8m + 32 - m^2 - m = 2 
-->  m^2 -7m - 30 = 0 
-->  (m - 10)(m + 3) = 0 
-->  m = 10

Option E

lacktutor · Answer

For any positive integer n, let f(n)=n(n+1) if n is even, and f(n)=n+3 if n is odd. If m is a positive integer such that 8f(m+1)−f(m)=2, then what is the value of m ?

Case1: m =ODD, m+1= EVEN
-->  8 (m+1)(m+1+1) - (m+3)=2 
 8(m+1)(m+2)- (m+3)=2 
 8m^{2} +24m+ 16 -m-3 -2=0 
 8 m^{2}+ 23m +11 =0 
 D =√529-352= √177 
--> The values of m are not integers.

Case2: m = EVEN, m+1 = ODD
-->  8(m+1+3)-m(m+1) =2  
 8(m+4 ) -m^{2}-m-2=0 
 m^{2} -7m -30=0 
 (m +3)(m -10)=0 
 m =-3  and  m=10 
-->  m =10  (positive integer)

Answer(E)

Archit3110 · Answer

starting with m as even
we have m+1 will be odd 
so 
8*(m+1+3)- (m)*(m+1) = 2
8m+32-m^2-m-2=0
m^2-7m-30=0
(m-10)*(m+3)=0
m=+10 and -3
IMO E

For any positive integer n, let f(n)=n(n+1) if n is even, and (n)=n+3 if n is odd. If m is a positive integer such that 8f(m+1)−f(m)=28f(m+1)−f(m)=2, then what is the value of m ?

A. 1
B. 3
C. 4
D. 6
D. 10

MrSengupta · Answer

I had a look at the comments , although they do suffice with the necessary solution , here's an easy technique to detemine .

If I first consider that m is odd , then (m+1) becomes even. If I solve this equation we get , 8m^2+23m+11=0 , we cannot have three postive terms to add to give a 0. So the odd possibility goes out .

Then is we plug and play with the options , we get that the answer is (E).

Crytiocanalyst · Answer

first we havve to figure out which case satisfies the given condition 
Anything multiplied by 8 is even and only when even number can lead to remainder as 2 
Therefore f(m) is even and f(m+1) odd since next digit of an even number is odd therefore 
8f(m+1)−f(m)=28f(m+1)−f(m)=2 
 8(m+4)−m(m+1)=28(m+4)−m(m+1)=2
 8m+32−m2−m=28m+32−m2−m=2
 m2−7m−30=0m2−7m−30=0
 (m−10)(m+3)=0(m−10)(m+3)=0
 m=10
hence IMO E

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For any positive integer n, let f(n) = n(n + 1) if n is even, and f(n)

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For any positive integer n, let f(n) = n(n + 1) if n is even, and f(n)

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