xx is a two digit integer, the first term of a sequence of c

ardz24 wrote:xx is a two digit integer, the first term of a sequence of consecutive odd positive integers. If the sum of xx and the last term equals 100, and the product of xx and the last term is a number whose four digits equal, in order, x-1, x-1, x-2, x-2, then what is the sum of the four digits?

A. 6
B. 10
C. 14
D. 18
E. 22

I'm confused how to set up the formulas here. Can any experts help?

Testing the answers could help here. The question states that the four digits are X-1,X-1,X-2 and X-2 and wants to know their sum.

The sum of these digits is 4X-6. Test the answer choices to see if any can be eliminated:

4X-6=6 > X=3
4X-6=10> X= 4 > Eliminate because the problem states that XX is odd meaning that X is odd
4X-6=14> X=5
4X-6=18>X=6 > Eliminate because it's not odd
4X-6=22>X=7

So answer choices A,C and E remain. The problem states that the sum of XX and the last term in the series equals 100. Answer C begins with 55 and the least last number in the series is 57, the sum of which exceeds 100. That eliminates C

Answer choice E is eliminated for the same reason, leaving only Answer A. Since the first term in that series is 33, the last term must be 67. You can prove that those numbers multiplied together provide the required 4 digit number, but it is not necessary to answer the problem

ardz24 wrote:xx is a two digit integer, the first term of a sequence of consecutive odd positive integers. If the sum of xx and the last term equals 100, and the product of xx and the last term is a number whose four digits equal, in order, x-1, x-1, x-2, x-2, then what is the sum of the four digits?

A. 6
B. 10
C. 14
D. 18
E. 22

I'm confused how to set up the formulas here. Can any experts help?

I like Regor's approach of using the answer choices and a bit of logic here.

Alternatively, we can think about it like this: first, notice that the sample space is fairly small. xx is an ODD number, So only 11, 33, 55, 77, and 99 are possibilities to start. Moreover, it's the first term in the sequence of consecutive odds. If the sum of this number and the last term is 100, then 55, 77, are all out, as those terms would all have to be larger than the last term. (55 + 45 = 100, but 45 can't be the last term if 55 is the first term, etc.)

That leaves us with two options: 11 or 33.

If the first term is 11, the last term would have to be 89 if the two must sum to 100. 11*89 will only be a 3-digit number.
Therefore, we know the first term has to be 33, and the last term 67. No need to do the math here. We know x = 3. So x - 1 = 2 and x - 2 =1, giving us a four digit number of x-1, x-1, x-2, x-2 = 2211. The sum of the four digits is 2 + 2 + 1 + 1 = 6. The answer is A