please help. thanks

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silvia928: Senior | Next Rank: 100 Posts; Posts: 37; Joined: Sat May 24, 2008 8:59 pm

by silvia928 » Wed Jun 04, 2008 3:29 pm

If n is a positive integer, what is the remainder when 3^(8n+3)+2 is divided by 5?

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

ANSWER IS....

(E)

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Source: Beat The GMAT — Problem Solving | Wed Jun 04, 2008 3:29 pm

ksh: Master | Next Rank: 500 Posts; Posts: 106; Joined: Sun Feb 17, 2008 10:09 pm; Thanked: 3 times

by ksh » Wed Jun 04, 2008 11:38 pm

Consider n=1
so, 3^11+2=(3^4)^2*3^3+2
so, last digits 1*7+2
=9 Now divide by 5, remainder=4
thats the answer E

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getneonow: Senior | Next Rank: 100 Posts; Posts: 42; Joined: Fri Mar 14, 2008 1:56 am; Location: India; Thanked: 4 times; GMAT Score:730

by getneonow » Thu Jun 05, 2008 12:19 am

By Eulers theorem 3^4 mod 5 = 1 .

so 3^8n mod 5 = 1

3^(8n+3) + 2 mod 5 = 3^8n mod5 * 27 mod5 + 2 mod5

= 1*2 + 2 = 4

Neo

MBA : My passion and My pursuit

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AleksandrM: Legendary Member; Posts: 566; Joined: Fri Jan 04, 2008 11:01 am; Location: Philadelphia; Thanked: 31 times; GMAT Score:640

by AleksandrM » Thu Jun 05, 2008 9:45 am

3 taken to a crazy power is one of GMAC favorites. The beauty of 3 is that it has a repeating pattern of the units digit: 3, 9, 7, 1, 3, 9....ad infinitum.

Therefore, if you write it out:

3^1 = 3

3^2 = 9

3^3 = 27 (units digit is 7)

3^4 = n1

3^5 = n3

...
...
3^11 = n7

7 + 2 = 9 and 9/5 (as has already been shown above) = remainder of 4.

There is a temptation to do a long division here, which will give you a 1.8, so you end up without a remainder since. However, do grade 3 long division where decimal places do not yet exist. So, you end up with a remainder of 4.

https://second-lap.blocked/

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beeparoo: Master | Next Rank: 500 Posts; Posts: 200; Joined: Sun Jun 17, 2007 10:46 am; Location: Canada; Thanked: 9 times

by beeparoo » Fri Jun 13, 2008 4:15 pm

Ummmm.....

While I thank AleksandrM for his super handy note about the trendy "Powers of 3", I'm curious as to why you guys decided to settle on a situation where n =1.

What if n = 2?

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beeparoo: Master | Next Rank: 500 Posts; Posts: 200; Joined: Sun Jun 17, 2007 10:46 am; Location: Canada; Thanked: 9 times

by beeparoo » Fri Jun 13, 2008 4:19 pm

Oh - nevermind. I see that when n = 2, the exponent is 19 and this also results in a ones-digit value of 7.

Add a 2, which equals 9 and et voila, a remainder of 4 all over again.

Without sweating the extra possibilities, n = 3, 4, 5.. etc. I suspect that all results will bear a ones-digit value of 9 each time.

BUT! Is there any (mathematical) way to know for sure??

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airan: Master | Next Rank: 500 Posts; Posts: 127; Joined: Tue May 13, 2008 12:28 am; Thanked: 2 times; GMAT Score:710

by airan » Sat Jun 14, 2008 6:56 am

Answer is in question.
8n+3, every time adding 3 to any multiple. So it will always be 3rd degree
like
n3
n9
n7
n1
and repeat ..so 8n+3 will always come to n7 category....add 2 it will become 9, Hence remainder is 4.

Thanks
Airan

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