Given M=1993^1997 + 1997^1993. What is the remainder of M

Question

Given M=1993^1997 + 1997^1993. What is the remainder of M divided by 15 ( means M/15)?
A. 11
B. 12
C. 13
D. 14
E. 15

FN · Answer

hey lexi..how are u? what schools did u apply to?

OK answer to your tuff question...

well i only look at the unit digit..1993<--3; for 1997<--7

3^1997, well the unit digit will be the same as that of 3^7=7

7^3=3, so we know the sum will have a unit digit 0...

so now if we divide 15 this remainder can be either 0 or 5 or 15...

the only answer choice that is there is 15..so 15 would be my answer..

hobbit · Answer

it is a lousy question.... and i'd say there is not correct answer....

first one anwer choice is realy a hoax.... 15 cannot be the remainder of anything divided by 15....

also fresinha has started in the right direction but made a mistake...

it is correct (but not always helpful) to look at unit digits...
it is not correct to reduce 1993^1997 to 3^7...

unit digit of exponents of 3 shows periodicity of 4... 3,9,7,1,3,9,7,1...
so you can see what is the remainder when dividing the exponent by 4: for both 1997 and 1993 this is 1.
so the unit digit of 1993^1997 is same as 3^1=3
similarly for exponents of 7... there is periodicity 7,9,3,1...
and again the unit digit of 1997^1993 is same as 7^1=7

so the unit digits of M is 0.

but here fresinha made a small mistake...
if the unit digits is 0, then the remainder when dividing by 15 is either 0,5, or 10. which leaves us with no correct answer.

i say that this question is either from umreliable source or that there is a typo.

FN · Answer

good catch...1997 mod 4, will lead to the right exponent.. and yes the answer choices should 0, 5 and 10.. I am a bit too rusty..need to get this old mind of mine back into shape

laxieqv · Answer

I double checked the question, there's no typo....saying that 15 is a remainder may not be sound but it's logical anyways, it's just another way to express a remainder of 0 (the reason i put it this way is to make sure we are alert enough)

@Freshinha: thank you for your concern

..I've not applied to any school yet ...I'm done with GMAT but i have to wait till i graduate from undergraduate first

..How about you?!!When will you retake GMAT?!!

hobbit · Answer

ok... suppose for now that 15 means 0....

what i did still doesn't prove M is divisible by 15..... it merely proves that M is divisible by 5. if the answer choices where 0,2,3,5,7 you couldn't answer the question by the logic used so far....
you should still have to show that M is divisible by 3 (if it is div isible by 3 and 5 then it is div. by 15):

1993 leaves reaminder 1 when divided by 3. all exponents of such numbers leave remainer 1 when divided by 3.
1997 leaves remainder 2 when divided by 3. odd exponenents of such numbers leaves 2 when divided by 3 (and even exponents leave remainder 1). so together we have remainder 0 when divided by 3. so M is divisible by 3. and together with the former logic, it is surely divisible by 15.

however - there is a simpler way, more direct, to show that M/15 is an integer. it involves noticing that 1995 is divisible by 15.
so we have M=(15k-2)^1997 + (15k+2)^1993 (k is some integer)...
now, when expanding this, most terms of the expansions are clearly multiples of 15, except 2 terms: -2^1997 (from the left side exponent) and 2^1993 (from the right hand side).

2^1993-2^1997 = 2^1993(1-2^4) = -15*2^1993
so these two terms combined are divisible by 15. and since all the other terms are divisible by 15, M must be divisible by 15 as well.

devilmirror · Answer

If M is an Integer, the set of the remainder of M divided by 15 could only be {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14} If I were to follow the question strictly, I would not choose 15 as the answer because 15 is not in the set. Btw, good question! :wink:

laxieqv · Answer

Thank all of you for trying this problem

Hobbit, I find your method very close to the original solution

. From this problem, one thing we should remember in order to promptly solve other similar problems is: we can always write (a+b)^n in the form of A*a+b^n or a^n+ B*b (in which A and B are integers and we don't even need to care exactly what they are) 15= 3*5 , the two factors have no common factors other than 1 so we can examine the problem in term of M/3 and M/5 We can easily find the unit digit of M which is 0, thus M is divisible by 5. 1993^1997= ( 1992+1)^1997= A*1992+1^1997. Notice that 1992 is divisble by 3 1997^1993= (1998-1)^1993= B*1998 -1^1993 . Notice that 1998 is divisible by 3. Add them up, 1^1997 and 1^1993 eliminate e/other. Thus, M= A*1992+B*1998 which is obviously divisible by 3. Finally, M is divisible by 3 and 5, thus divisible by 15.

Given M=1993^1997 + 1997^1993. What is the remainder of M

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