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Comment on Remainder when Divided by 7
My approach was similar to
Great approach!
Great approach!
My approach is like this.
Once we equate both equations of x, we get 5q = 7k. Once looking at this, we will get 35 in our mind where 5 and 7 meets. So q=7 and k=5.
Then, q/7 means 7/7 which leaves remainder 0.
Great approach.
Great approach.
https://gmatclub.com/forum
here sir i did this
unit digit of 12^12= 6
unit digit of 13^13 = 3
unit digit of 12^12 +13^13 will be 9
and unit digit of (14^14*15^15) will be 0
but 14^14*15^15 > 12^12+13^13
when we subtract two numbers last digit would be 10-9 = 1
eg 20-9 last digit will be 1 30-9 will be 1
sir is this approach correct?
That's a perfectly valid
That's a perfectly valid solution. Nice work!
Cheers,
Brent
dear Brent what will happen
Great question! Let's find
Great question! Let's find out what happens.
GIVEN: When positive integer x is divided by 5, the quotient is q and the remainder is 1.
If x = 1, then q = 0 (since 5 divides into 1 ZERO times with remainder 1)
QUESTION: What is the remainder when q is divided by 7?
Since q = 0, the remainder will be 0, when 0 is divided by 7.
Answer: A
Cheers,
Brent
Hi Brent,
Could you please explain the following question? https://gmatclub.com/forum/m17-184114.html#p1414882
Thank you! :)
Tricky!!!
Tricky!!!
Here's my full solution: https://gmatclub.com/forum/m17-184114.html#p2693156
Hi Brent,
Can you please explain this part in more details?
If 10^m − 1 is divisible by 3, then 10^m is 1 greater than a multiple of 3
In other words 10^m = 3j + 1 for some integer j.
At this point, we can see that 10^m + n = (3j+1) + (3k+2) = 3j + 3k + 3 = 3(j+k+1)
Since we can be certain that 3(j+k+1) is a multiple of 3, we know that 3(j+k+1) divided by 3 must leave a remainder of 0
Also, Happy New Year!
If 10^m − 1 is divisible by 3
If 10^m − 1 is divisible by 3, we can say: 10^m − 1 = 3j (for some integer j)
Now take 10^m − 1 = 3j and add one to both sides to get: 10^m = 3j + 1
Earlier in my solution, I showed that n = 3k + 2
So we can now write: 10^m + n = (3j+1) + (3k+2) = 3j + 3k + 3 = 3(j+k+1) etc
ASIDE: I see that, in part of my solution, the variable m turned into the variable b. I've edited that part on GMAT Club.
Happy New Year to you!!!
Makes sense, thank you so
https://gmatclub.com/forum/if
Can I have the solution for this?
How do I tackle questions like this? I forgot.
Here's how I would solve the
Here's how I would solve the question on test day: https://gmatclub.com/forum/if-n-is-an-integer-greater-than-6-which-of-th...
wait it says n is an integer
Great question! I should have
Great question! I should have explained myself.
I went back and added an explanation about why it's okay to test numbers smaller than 6.
Hi Brent I used a different
X=5Q+1
X=7K+1
5Q+1=7K+1
5Q+7K=0
Then Q must be zero 0/7=0 with zero remainder
Is that a correct approach?
That approach is similar to
That approach is similar to the first solution I provide.
You made just one small mistake however.
Once we have: 5Q + 1 = 7K + 1
We can then say: 5Q = 7K, which means 5Q - 7K = 0 (not 5Q + 7K = 0, as you have in your solution)