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AMC 10 USA Math Olympiad

Combination of Equations | SMO, 2010 | Problem No. 7

Try this beautiful problem from Singapore Mathematical Olympiad, SMO, 2010 – Problem 7 based on the combination of equations.

Try this beautiful problem from Singapore Mathematical Olympiad, SMO, 2010 – Problem 7 based on the combination of equations.

Problem – Combination of Equations (SMO Entrance)


Find the sum of all the positive integers p such that the expression (x-p) (x – 13) + 4 can be expressed in the form (x+q) (x+r) for distinct integers q and r.

  • 26
  • 27
  • 16
  • 20

Key Concepts


Basic Algebra

Combination of Terms

Generator of a group

Check the Answer


But try the problem first…

Answer: 26

Source
Suggested Reading

Singapore Mathematical Olympiad

Challenges and Thrills – Pre College Mathematics

Try with Hints


First hint

If you got stuck in this problem start this problem using this hint :

Start with the given hint

(x-p) (x-13) +4 = (x+q)(x+r)

Let’s try to minimize the expression by taking x= -q

so , (-q -p)(-q -13) = -4 , which becomes (q+p) (q+13) = -4

As it is already given p and q are integers we can come up with many cases .

Try to find out the different cases we can have ………………………..

Second Hint

Starting after the last hint :

p+q = 4 and q+13 = -1 ………………………………..(1)

q+p = -4 and q +13 = 1……………..(2)

p+q = 2 and q+13 = -2 and …………………………(3)

p+q = -2 and q+13 = 2 …………………………(4)

For 1st case its simple calculation that we get q = -14 and p = 8

The initial expression becomes (x-p) (x-13) +4 = (x-14) (x-17)

For 2 nd case :

q= -12 and p = 8

so the initial expression becomes : (x-p)(x-13)+4 = (x-9)(x-12)

try the rest of the cases…………

Final Step

Now let’s talk about 3rd case ,

q= -15 , p = 17 and

hence (x – p) (x – 13) +4 = \(( x – 15)^2\) which is not true to this problem .

For last case , we obtain q = -11, q = 9 so the initial

(x- p) (x-13) +4 = \((x-11)^2\)

So p is 8 and 18 which add upto 8 +18 = 26 -(Answer)

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