Week 26

Example 1. How many three-digit numbers are there such that the number is 19 times the sum of their digits?

Example 2. Find all real polynomials f(x) such that f(x − 1)f(x + 1) = f(f(x)) for all xR.

Of course, you can find the solution just below, but it is highly recommended that you first try to solve it on your own.

Just remember the words of Paul Halmos, who says “the only way to learn mathematics is to do mathematics”.

Solution 1. Let the three-digit number be (100a + 10b + c), where a, b and c are integers between 0 and 9, and a > 0 (as it is a three-digit number).

Here, sum of the digits = a + b + c.

The condition that the number is 19 times the sum of their digits can be written as

(100a + 10b + c) =19(a + b + c)

Or, 81a − 9b −18c = 0

Or, 9ab − 2c = 0

Or, b = 9a − 2c.

This is the required relation between a, b and c. Moreover, 0 < a < 9 and b, c can take values from 0 to 9. Following combinations are possible:

  • b = 0 when a = 2 and c = 9. The number is 209.
  • b = 1 when a = 1 and c = 4. The number is 114.
  • b = 2 when a = 2 and c = 8. The number is 228.
  • b = 3 when a = 1 and c = 3. The number is 133.
  • b = 4 when a = 2 and c = 7. The number is 247.
  • b = 5 when a = 1 and c = 2. The number is 152.
  • b = 6 when a = 2 and c = 6. The number is 266.
  • b = 7 when a = 1 and c = 1. The number is 171.
  • b = 8 when a = 2 and c = 5. The number is 285.
  • b = 9 when a = 1 and c = 0 or a = 3 and c = 9. The numbers are 190 and 399.

Therefore, the numbers are 114, 133, 152, 171, 190, 209, 228, 247, 266, 285 and 399.

Solution 2. Let d = deg f. We know that (fg)(x)= f(x)g(x) and (f g) = f(g(x)).

Hence deg (fg) = deg f + deg g and deg (f g) = deg f × deg g.

Then LHS of the equation is of degree 2d, whereas the RHS is of degree d2. Then 2d = d2, implying that d = 0 or d = 2.

A polynomial of zero degree is constant, and the only constant f(x) = c satisfying this equation is c = 0 or c = 1. Let us choose a degree 2 polynomial as f(x) = ax2 + bx + c, where a ≠ 0. Then

f(x − 1) f(x + 1) = f(f(x)) = a{f(x)}2 + bf(x) + c

⇒ {a(x − 1)2 + b(x − 1) + c}{a(x + 1)2 + b(x + 1) + c}  = a(ax2 + bx + c)2 + b(ax2 + bx + c) + c

⇒ {a(x2 – 2x + 1) + b(x − 1) + c}{a(x2 + 2x + 1) + b(x + 1) + c}  = a(a2x4 + b2x2 + c2 + 2abx3 + 2bcx + 2cax2) + b(ax2 + bx + c) + c

⇒ {ax2 + (b – 2a)x + (ab + c)} {ax2 + (b + 2a)x + (a + b + c)} = a{a2x4 + 2abx3 + (b2 + 2ca) x2 + 2bcx  + c2} + bax2 + b2xbc + c

a2x4 + {(b + 2a)a + (b – 2a)a}x3 + {a(a + b + c) + a(ab + c) + (b2 – 4a2)}x2 + {(b − 2a) (a + b + c) + (b +2a) (ab + c)}x + (ab + c) (a + b + c) = a3x4 + 2a2bx3 + {a(b2 + 2ca) + ab}x2 + {2abc + b2}x  +(ac2 + bc + c)

Equating the coefficients of like powers of  on both sides, we get

  • a2 = a3   or, a(1 − a) = 0   or, a = 1 since a = 0 is not possible;
  • (1 + b + c) + (1 – b + c) + (b2 – 4) = (b2 + 2c) + b   or, b = – 2;
  • (ab + c) (a + b + c) = ac2 + bc + c      or, c = 1.

Hence f(x) = x2 − 2x + 1 = (x − 1)2.

It is easy to verify that, this is indeed the solution.

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