If P = m^3 - m, where m and P are positive integers, and m is given by

Question

If  P = m^3 - m , where m and P are positive integers, and m is given by the expression  m=(10x)^n+5^4 , where x and n are positive integers, then what is the remainder when P is divided by 4 ?

A. 4
B. 3 
C. 2
D. 1 
E. 0

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chetan2u · Answer

P = m^3 - m=m(m^2-1)=(m-1)m(m+1) .
Thus, P is a multiple of 3 consecutive numbers. So surely a multiple of 6 and an even number. Answer is either 2 or 0.

If m is odd, then both m-1 and m+1 are even, and thus P is a multiple of at least 8.
If m is even, the answer will depend only on m.

Now m =  m=(10x)^n+5^4  = EVEN + ODD = ODD

Thus, P is a multiple of at least 8.

E

Crytiocanalyst · Answer

let assume n=1 and decide the value of x later ,

then m=10x +5^4 .............1

when m^3 is divided by 4 and we assume n=2 so that the remainder of the first term is zero and the second term will yiels one

THerefore 1^3 - 1 =0

THerefore IMO E

mysterymanrog · Answer

Realise that 10x^n will be just some number ending in 1...00 for x>=2 or 10 for x=1.

For the case of x=1, the last two digits of the 3 expanded numbers of p become
(25+10-1)(25+10)(25+10+1)
(34)(35)(36)
36 is divisble by 4, so all of P is divisible by 4.

No need to do the case of x>=2, but in this case the last two digits become: 
(24)(25)(26), which have enough 2s to be divided by 4.

E.

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SeedWasp · Answer

P can be re-written as m^3-m=m(m^2-1)=(m-1)m(m+1). 
The number of m will be ending in 5 regardless of what are the values of x and n. This makes (m-1) and (m+1) even numbers. 
Thus this number will be completely divisible by 4 (since it is a product of 2 even numbers)

If P = m^3 - m, where m and P are positive integers, and m is given by

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If P = m^3 - m, where m and P are positive integers, and m is given by

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