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by eaakbari » Sat Apr 17, 2010 10:46 pm
Asumming your question was p^3-p^2 and not 3p-2p

Since units of p^3 -P^2 = 0

Units digit of p^3 and p^2 is the same

It can be one of the following
2
4
6
8

As these are even and positive

Out of all only 6 satisfies condition.

Hence units p = 6
units p +3
= 9
IMO 9
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by pradeepkaushal9518 » Sat Apr 17, 2010 11:12 pm
sir why not 10 which also satisfy the condition

10^3=1000
10^2=100
1000-100=900

that is main question i think?
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by eaakbari » Sat Apr 17, 2010 11:36 pm
pradeepkaushal9518 wrote:sir why not 10 which also satisfy the condition

10^3=1000
10^2=100
1000-100=900

that is main question i think?
Kya Sir bula raha hain yaar,
Given that p is a positive even integer with a positive units digit
Since units digit is a positive integer, 0 is neither positive nor negative. Hence it cannot be 0.

HTH
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by pradeepkaushal9518 » Sun Apr 18, 2010 4:03 am
ok now it is clear. do i know you ?
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by eaakbari » Sun Apr 18, 2010 4:05 am
pradeepkaushal9518 wrote:ok now it is clear. do i know you ?
Well i guess not, unless you know Ebrahim Akbari.
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