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by dtweah » Thu May 14, 2009 3:48 pm
Let N denote the smallest positive integer that satisfies the two conditions:
(i) If N is divided by 11 then the remainder is 9
(ii) If N is divided by 7 then the remainder is 5.
The sum of the digits of N is

(a) 10
(b) 11
(c) 12
(d) 13
(e) 14
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by cramya » Thu May 14, 2009 3:59 pm
C 12

75

So the numbers would be 75+77k wherek=0,1,2,3,4....

I had to check 11's multiples and add 9 and check. It dint take me too long though.

There might be an easier way

Regards,
CR
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by artistocrat » Thu May 14, 2009 8:25 pm
I solved this the old fashioned way, but what is an easier method than setting up a list of numbers that satisfies both conditions?
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by avenus » Fri May 15, 2009 2:53 am
artistocrat wrote:I solved this the old fashioned way, but what is an easier method than setting up a list of numbers that satisfies both conditions?
Maybe this:

11m + 9 = 7p + 5. This is what we're told. Add 2 to both sides of the equation:
11m + 11 = 7p + 7
11(m+1) = 7(p+1)
So
m+1 = 7, p+1 = 11
m = 6, p = 10
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