AMC 10 Step by Step

2011 AMC 10A Problem 4

P1-10~2 minPrint

Let XX and YY be the following sums of arithmetic sequences: X=10+12+14+⋯+100,Y=12+14+16+⋯+102.\begin{array}{rcl} X &=& 10 + 12 + 14 + \cdots + 100, \\ Y &=& 12 + 14 + 16 + \cdots + 102. \end{array} What is the value of Y−XY - X ?

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Key insight

The two sums share every even number from 12 to 100. Those cancel when you subtract, leaving just 102 minus 10.

Solution

Adding up each sum would take a while. Since we only want the difference, look for what the two sums have in common.

1. Write out the start and end of each sum:
- X=10+12+14+⋯+98+100X = 10 + 12 + 14 + \cdots + 98 + 100
- Y=10+12+14+⋯+98+100+102Y = \phantom{10 + {}}12 + 14 + \cdots + 98 + 100 + 102
2. Both sums contain every even number from 1212 up to 100100.
3. The only differences: XX also has a 1010 at the start, and YY also has a 102102 at the end.
4. When you subtract, all the shared numbers cancel. What is left is
Y−X=102−10=92.Y - X = 102 - 10 = 92.

The answer is (A) 92\boxed{\textbf{(A)}\ 92}.

Why this works

When two sums are almost the same, don't add up either one. Line them up, cross out what they share, and subtract only what is different. The same "cancel the overlap" idea shows up in many harder sum problems.

Alternative approach

Match the terms in order: 1212 with 1010, 1414 with 1212, and so on, up to 102102 with 100100. Each term of YY is 22 more than the matching term of XX.

Count the terms of XX: from 1010 to 100100 in steps of 22 there are 100−102+1=46\frac{100 - 10}{2} + 1 = 46 numbers. So YY is ahead by 22 a total of 4646 times, and Y−X=46×2=92Y - X = 46 \times 2 = 92.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Both sums run through the even numbers 12,14,…,10012, 14, \dots, 100. Beyond that, XX has the extra term 1010 at the front and YY has the extra term 102102 at the end.

When we subtract, every shared term cancels:
Y−X=102−10=92.Y - X = 102 - 10 = 92.

The answer is (A) 92\boxed{\textbf{(A)}\ 92}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.