What will be the remainder when 13^7 + 14^7 + 15^7 + 16^7 is divided by 58?
57
1
30
0
28
The answer is D) 0, but how do you reach the solution?
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- Luci
- Harish Dorai
I am not sure about the rule that you mentioned. This is how I approached and I narrowed it down to 3 choices now (C), (D) and (E). This is what I did (it is a little bit lengthy). Even though it looks lengthy and complicated on paper, it doesn't take much time to think through this.
Step 1: Find the units digit of 13^7 + 14^7 + 15^7 + 16^7.
For 13^7 the units digit is 7, for 14^7 the units digit is 4, for 15^7 the units digit is 5 and for 16^7 it is 6.
7 + 4 + 5 + 6 = 22, which means the units digit of 13^7 + 14^7 + 15^7 + 16^7 is 2.
Step 2:
Let us assume the the above sum is some big number, XYZ.......2 (the last digit is 2 as we figured out in Step 2). And we will see what could be the remainders by playing with the answer choices.
If you examine answer choice (A), it says the reminder is 57. In order for us to get a remainder 57, for a number with units digit 2 divided by 58, we should have a multiple that ends in 5 for the number 58. I will explain it using some small numbers.
If you divide 352 by 59, we will get a remainder 57. The calculation is depicted below.
5
------------
59) 3 5 2
2 9 5
--------
5 7
The very reason why we got a 57 above (or a remainder with units digit 7) is because we have a multiple for 59 that ends in 5 which is 295 which when subtracted from 352 yields the remainder 57.
Now coming back to our question, there is no multiple for 58 that ends in 5. So 57 cannot be a remainder. So we can eliminate answer choice (A).
Similarly we can eliminate answer choice (B) also. In order for us to get a remainder 1, we should have a multiple for 58 that ends in 1. We know there is no multiple for 58 that ends in 1.
Like above we cannot really eliminate the remainders 28, 30 and 0. So I am stuck here.
Step 1: Find the units digit of 13^7 + 14^7 + 15^7 + 16^7.
For 13^7 the units digit is 7, for 14^7 the units digit is 4, for 15^7 the units digit is 5 and for 16^7 it is 6.
7 + 4 + 5 + 6 = 22, which means the units digit of 13^7 + 14^7 + 15^7 + 16^7 is 2.
Step 2:
Let us assume the the above sum is some big number, XYZ.......2 (the last digit is 2 as we figured out in Step 2). And we will see what could be the remainders by playing with the answer choices.
If you examine answer choice (A), it says the reminder is 57. In order for us to get a remainder 57, for a number with units digit 2 divided by 58, we should have a multiple that ends in 5 for the number 58. I will explain it using some small numbers.
If you divide 352 by 59, we will get a remainder 57. The calculation is depicted below.
5
------------
59) 3 5 2
2 9 5
--------
5 7
The very reason why we got a 57 above (or a remainder with units digit 7) is because we have a multiple for 59 that ends in 5 which is 295 which when subtracted from 352 yields the remainder 57.
Now coming back to our question, there is no multiple for 58 that ends in 5. So 57 cannot be a remainder. So we can eliminate answer choice (A).
Similarly we can eliminate answer choice (B) also. In order for us to get a remainder 1, we should have a multiple for 58 that ends in 1. We know there is no multiple for 58 that ends in 1.
Like above we cannot really eliminate the remainders 28, 30 and 0. So I am stuck here.
- StaceyKoprince
- ManhattanGMAT Staff
- Posts: 9360
- Joined: Wed Oct 19, 2005 9:05 am
- Location: Montreal
- Luci
Not much, this is the website: http://www.4gmat.com/ and there they have a Free GMAT math test with 10 questions.
After completing the test they tell you the answers but dont explain them.
I had doubts with that particular answer
Thanks
After completing the test they tell you the answers but dont explain them.
I had doubts with that particular answer
Thanks
- StaceyKoprince
- ManhattanGMAT Staff
- Posts: 9360
- Joined: Wed Oct 19, 2005 9:05 am
- Location: Montreal
This is an... odd... one. I've never seen the test require us to know this particular theorem. I'm not sure I'd use this source to study.
58 = 2*29
I need to figure out the remainder when the sum is divisible by 2 and also by 29.
13^7 + 14^7 + 15^7 + 16^7
odd + even + odd + even = even, so this is divisible by 2 with a remainder of zero.
This part will be harder to follow. When evaluating remainders when dividing by 29, 15 can also be written as -14 and 16 can also be written as -13 (because 29 + (-14) = 15 and 29 + (-13) = 16). This lets us re-write the expression:
13^7 + 14^7 + (-14)^7 + (-13)^7
The middle two terms cancel and the outer two terms cancel, leaving a sum of zero. Zero divided by 29 gives a remainder of zero.
Overall, then, I have a remainder of zero. Again, I have NEVER (in more than ten years) seen a test question that required this knowledge so I really wouldn't worry about this.
58 = 2*29
I need to figure out the remainder when the sum is divisible by 2 and also by 29.
13^7 + 14^7 + 15^7 + 16^7
odd + even + odd + even = even, so this is divisible by 2 with a remainder of zero.
This part will be harder to follow. When evaluating remainders when dividing by 29, 15 can also be written as -14 and 16 can also be written as -13 (because 29 + (-14) = 15 and 29 + (-13) = 16). This lets us re-write the expression:
13^7 + 14^7 + (-14)^7 + (-13)^7
The middle two terms cancel and the outer two terms cancel, leaving a sum of zero. Zero divided by 29 gives a remainder of zero.
Overall, then, I have a remainder of zero. Again, I have NEVER (in more than ten years) seen a test question that required this knowledge so I really wouldn't worry about this.
Stacey Koprince
Instructor
Director, Content & Curriculum
ManhattanPrep
Instructor
Director, Content & Curriculum
ManhattanPrep
- shaji
The concept here is the 'Remainder Theorem' and this problem is a 15seconds job.
Well!!! good the GMAT hasn't got into the habbit of 'surprises'. If they do take a fancy to such stuff ,perhaps, the tests would be more 'challenging'
Well!!! good the GMAT hasn't got into the habbit of 'surprises'. If they do take a fancy to such stuff ,perhaps, the tests would be more 'challenging'
skoprince Wrote:This is an... odd... one. I've never seen the test require us to know this particular theorem. I'm not sure I'd use this source to study.
58 = 2*29
I need to figure out the remainder when the sum is divisible by 2 and also by 29.
13^7 + 14^7 + 15^7 + 16^7
odd + even + odd + even = even, so this is divisible by 2 with a remainder of zero.
This part will be harder to follow. When evaluating remainders when dividing by 29, 15 can also be written as -14 and 16 can also be written as -13 (because 29 + (-14) = 15 and 29 + (-13) = 16). This lets us re-write the expression:
13^7 + 14^7 + (-14)^7 + (-13)^7
The middle two terms cancel and the outer two terms cancel, leaving a sum of zero. Zero divided by 29 gives a remainder of zero.
Overall, then, I have a remainder of zero. Again, I have NEVER (in more than ten years) seen a test question that required this knowledge so I really wouldn't worry about this.
- JAMGAJR
Remainder
Another way
(15^7 + 14^7) + (13^7 + 16^7).......... remember that a^n + b^n is always divisible by a+b when n is odd
so both of the expressions are divisible by 29
so 1/2* ( odd + even + odd + even)= 1/2*(even)= divisible
remainder 0
(15^7 + 14^7) + (13^7 + 16^7).......... remember that a^n + b^n is always divisible by a+b when n is odd
so both of the expressions are divisible by 29
so 1/2* ( odd + even + odd + even)= 1/2*(even)= divisible
remainder 0
- StaceyKoprince
- ManhattanGMAT Staff
- Posts: 9360
- Joined: Wed Oct 19, 2005 9:05 am
- Location: Montreal
Well!!! good the GMAT hasn't got into the habbit of 'surprises'. If they do take a fancy to such stuff ,perhaps, the tests would be more 'challenging'
Just note that this question didn't come from GMATPrep or OG - it came from some other source - so this has no bearing on whether the official GMAT may or will use such concepts on the real test.
Stacey Koprince
Instructor
Director, Content & Curriculum
ManhattanPrep
Instructor
Director, Content & Curriculum
ManhattanPrep
- anadi
here is a solution
That might make it look more like a gmat question and concept
(a-b)^2 = a^2+b^2-2ab
(a-b)^3 = a^3-b^3+3ab^2-3ba^2
If we expand any power of a+b like this, there will be one and only one part of multinomial that will have either a or b in their highest power.
now,
13^7+14^7+15^7+16^7 can be written as,
13^7+14^7+(29-14)^7+(29-13)^7, when we expand the later 2 parts, (-14)^7 and (-13)^7 wil be the only once that will not have 29(or a power of 29) multiplied to them. So 13^7+14^7+(-14)^7+(-13)^7 = 0.
Rest of the expanded part will have a 29(or a power of 29) as a factor in each, and from the original 13^7+14^7+15^7+16^7, we know that this must be even. Hence we got 2 and 29 as factors of equation and that proves this as well as the remainder theorem.
This applies to only odd powers since even powers of -14 and -29 would not have cancelled out with the positive ones.
(a-b)^2 = a^2+b^2-2ab
(a-b)^3 = a^3-b^3+3ab^2-3ba^2
If we expand any power of a+b like this, there will be one and only one part of multinomial that will have either a or b in their highest power.
now,
13^7+14^7+15^7+16^7 can be written as,
13^7+14^7+(29-14)^7+(29-13)^7, when we expand the later 2 parts, (-14)^7 and (-13)^7 wil be the only once that will not have 29(or a power of 29) multiplied to them. So 13^7+14^7+(-14)^7+(-13)^7 = 0.
Rest of the expanded part will have a 29(or a power of 29) as a factor in each, and from the original 13^7+14^7+15^7+16^7, we know that this must be even. Hence we got 2 and 29 as factors of equation and that proves this as well as the remainder theorem.
This applies to only odd powers since even powers of -14 and -29 would not have cancelled out with the positive ones.
13 posts Page 1 of 1