challenge questions  TOPIC_SOLVED

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challenge questions

Postby nazz on Tue Oct 27, 2009 2:59 pm

1.
A number has three digits and is equal to 12 times the sum of its digits what the number is?

Solution:

Assume three digit number = xyz, then

100x + 10y + z = 12(x + y + Z)

100x + 10y + z = 12x + 12y + 12Z

Solve to get

88x = 2y + 11z

Take x=1, then y=0 and z= 8 in order to satisfy the equation.

Three digit number is 108, hence 108 = 12(1+0+8).

Is there any other way to solve this problem?

2.

The sum of first N integers is a three digit number with all of its digits equal, what is the value of N?

1,2,3,4,5,6,7……………. Is an arithmetic series

S= n/2[2a + (n-1) d] = n (n +1) / 2

n (n +1) / 2 = 111x

Since sum of first integers is a three digit number with all digits equal
Hence it xxx = 100x +10x +x = 111x.

By trial and taking x=6, and solving for n, n2 +n – 1332 = 0, I will get n=36, is there any other way of solving;
n (n +1) / 2 = mod 111, and getting value of n=36 from this equation?
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  TOPIC_SOLVED

Postby stapel_eliz on Tue Oct 27, 2009 10:03 pm

nazz wrote:Is there any other way to solve this problem?

They've given you information for one equation in terms of three variables. As such, guess-n-check and using the fact that the digits must be whole numbers between 0 and 9 is pretty much the only way to go, unless you're studying Diophantine equations in graduate school. :shock:

nazz wrote:The sum of first N integers is a three digit number with all of its digits equal, what is the value of N?

To some extent, you may be able to simplify your work by finding the numbers of terms which give you no more than a two-digit number, and no less than a four-digit number, and establish some parameters for the value of n. But I don't know that this would, in practice, actually help.

Taking your quadratic, , plugging values in for , and seeing if you can solve the resulting equation for whole-number values of n is likely the best method. :wink:

Unfortunately, some parts of math can be just painful!
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