A number theory problem by Abhyuday Singh
How many integers between 11 and 1111 leave a remainder of 6 when divided by 9 and a remainder of 12 when divided by 21?
The answer is 18.
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3 solutions
I agree that the question seemed to be asking for two values. I've made the appropriate edits, knowing now that the answer is in fact 18.
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Thanks. I see that this problem has been edited.
Hi Brian, as a moderator, please do not edit problem significantly, because that doesn't update it for other who already answered the problem. Instead, simply report the problem and staff will handle it delicately.
Thanks for helping us make Brilliant better :)
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Sorry about that! Normally I don't hesitate to use the reporting system, but this time I both slipped up and overreached. I also didn't give Abhuyday the chance to learn how to edit his own problem for clarity. Thanks for the feedback. :)
Let N = 9 k 1 + 6 and N = 2 1 k 2 + 1 2
Comparing 2 1 k 2 + 6 = 9 k 1 + 6 = > 7 k 2 + 2 = 3 k 1 .
So, 3 ∣ 7 k 2 + 2 = > 7 ∣ k 2 + 2
The values of k 2 will be 1 , 4 , 7 , 1 0 , . . . .
The upper limit is 2 1 k 2 + ! 2 < 1 1 1 1 = > k 2 ≤ 5 2
So, required values of k 2 are ( 1 , 4 , 7 , 1 0 , . . . . . . , 4 9 , 5 2 )
There are 1 8 k 2 ′ s are there and each k 2 will corresponds to a unique integer N
So, the answer is 1 8
the least number which does so is 33 and the gcd of the 2 remainders is 63 so next numbers will follow an arithmetic progression of common difference 63 as the 1st number is 33 next numbers will be 96,159 and so on and the total numbers between 11 and 1111 will be 18
Could you rephrase the question so the answer to be input in the box is 18? It looks like one has to answer both 33 and 18.