What is the remainder when positive integer p is divided by 10?

The answer is D. Previously chose E, because I thought 7/10 has no remainder, but thanks to Nik18, I realized the remainder is still 7, so D for me.

Hi rulingbear,

if p=7 then the remainder will be 7 when divided by 10, because if Divisor > Dividend, then the remainder is Dividend itself. So D should be the answer

D is the answer.
But P can take any integer value whose unit digit is 7 ( P=7,17,27,37....).

Correct me if I'm wrong.

Wonderful question.

I started with Statement 2 first just for convenience:

Statement 1:
Similar concept -
p^11 + 11^p
11 raised to power of any number always has units digit of 1.
Therefore P^11 has to have units digit of 3 - in order for the sum to have units digit of 4 so as to leave 4 remainder when divided by 10
and only 7 satisfies that condition.
Hence, sufficient.

Statement 2: If the cube is leaving a remainder of 3 with 10, the units digit has to be 3.
Only the numbers with units digit 7 have a cube of units digit 3.
Hence p is xxx7, leaving remainder 7 with 10.
Sufficient

Answer is D.

Please kudos if you like the post.

Hi Niks,

Thanks for the reply. You are right, the remainder of 7/10 is 7, i.e [fraction]0[/fraction]10+7. It should be D.

Both 3 and 7 have the number three in the units digit. Are we (a) supposed to know this and (b) have the time to compute the cyclicality of both numbers, just to answer part of the question?

a) We aren't supposed to know this. We can check it in 10- seconds.
2,4,6,8 raised to any power can't give us 3 (as they are all even). Nor can 0,1,5, (they end in 0,1, and 5 respectively) and 9 (end in 1 or 9). Only option remain are 3 and 7. 3 gives in first power. and 7 gives in 3rd power. (3rd is not a high power to try)

b) Yes we have time to compute cyclicity:
we care about units digit only, so multiplying the digit with multiplier and taking only units digit is sufficient.
here, for 3 --> 3, 9 (units digit of 3X3), 7 (units digit of 9X3), 1 (units digit of 7X3), 3 we stop here.
we have 3^11. So just count it from left to right from 1 to 11. We arrive at 7.
Similarly for 7 ---> 7,9 (7X7), 3 (9X7), 1 (3X7), 7 stop here.
counting the same way will give you 3.
This process will take hardly 15 seconds.

So effectively, we'll need around 30 seconds to work this out. Maximum 45 if some of us are slow pickers...

I suppose I answered your query

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