Mathematical Enigma - II

Algebra Level 4

Let P ( x ) P(x) be a polynomial with non-negative integer coefficients. If P ( 1 ) = 6 P(1) = 6 and P ( 5 ) = 426 P(5) = 426 , find P ( 3 ) P(3) .


The answer is 100.

View wiki
Swapnil Das by Swapnil Das , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

5 4 = 625 > 426. So the degree P(x) can not be 4. P ( 5 ) = 426 , a n d 5 3 = 125 , 5 2 = 25 , 5 1 = 5 5^4=625>426.\text{ So the degree P(x) can not be 4.}\\ P(5)=426, \ and\ 5^3=125,\ 5^2=25,\ 5^1=5 , so the constant must be at least 1. But the constant term can be 1 or 6 only. Since P(1)=6, the constant term is only 1. Now in P(5) we are left with 425 with remaining coefficient add to 5. 5 ∗ 5 2 consumes all coefficients but P(5)=126 only . S o o n l y 5 3 s u p p l i e s t h e b u l k o f 425. 3 ∗ 5 3 m u s t s u p p l y 375. Remaining 50 is supplied by 2 ∗ 5 2 . L e t u s s e e w h a t w e h a v e g o t . P ( x ) = 3 x 3 + 2 x 2 + 1. P ( 5 ) = 3 ∗ 5 3 + 2 ∗ 5 2 + 1 = 426 , a n d P ( 1 ) = 6 , S o P ( 3 ) = 3 ∗ 3 3 + 2 ∗ 3 2 + 1 = 100 \text{But the constant term can be 1 or 6 only. Since P(1)=6, the constant term is only 1. } \\ \text{Now in P(5) we are left with 425 with remaining coefficient add to 5.}\\ 5*5^2 \text{consumes all coefficients but P(5)=126 only}. \\ So\ only\ 5^3 \ supplies\ the\ bulk\ of\ 425.~~~ 3*5^3\ must\ supply\ 375.\\ \text{Remaining 50 is supplied by } 2*5^2.\\ Let\ us \ see\ what\ we\ have\ got.\ \ \ P(x)=3x^3+2x^2+1.\\ P(5)=3*5^3+2*5^2+1=426,\ and\ P(1)=6,\\ So\ \ P(3)=3*3^3+2*3^2+1=\Large\ \ \ \ \ \color{#D61F06}{100}

7 Helpful 0 Interesting 0 Brilliant 0 Confused
Kenny Lau
Oct 16, 2015

The coefficients sum up to be 6 6 , because P ( 1 ) = 6 P(1)=6 .

Since P ( 5 ) P(5) would be a 0 ( 5 0 ) + a 1 ( 5 1 ) + a 2 ( 5 2 ) + a 3 ( 5 3 ) + ⋯ a_0(5^0)+a_1(5^1)+a_2(5^2)+a_3(5^3)+\cdots , let us express 426 426 in base-5.

42 6 10 = 320 1 5 426_{10}=3201_{5} , which satisfy that the digit sum is 6 6 .

Therefore P ( x ) = 3 x 3 + 2 x 2 + 1 P(x)=3x^3+2x^2+1 .

4 Helpful 0 Interesting 0 Brilliant 0 Confused

For completeness, you need to justify why there is only that one solution. In particular, it is tricky (not as straightforward as you think it is) because P ( 1 ) > 5 P(1) > 5 .

See my threaded comment for why it is incomplete.

Calvin Lin Staff - 5 years, 7 months ago

Log in to reply

Why is it tricky ? I still don't get why his solution is incomplete ?

Krutarth Patel - 2 years, 7 months ago

Log in to reply

The incomplete step is going from just P ( 5 ) = 426 = a 0 × 5 0 + a 1 × 5 1 + … ⇒ P ( x ) = a 0 + a 1 x + … P(5) = 426 = a_0 \times 5^0 + a_1 \times 5^1 + \ldots \Rightarrow P(x) = a_0 + a_1x + \ldots . This is mainly because the representation is not unique , and so for each representation, that gives us a possible polynomial to try. For example, P ( 5 ) = 426 = 6 × 5 0 + 4 × 5 1 + 1 × 5 2 + 3 × 5 3 P(5) = 426 = 6 \times 5^ 0 + 4 \times 5 ^ 1 + 1 \times 5^2 + 3 \times 5^3 , but P ( x ) ≠ 6 + 4 x + 1 x 2 + 3 x 3 P(x) \neq 6 + 4x + 1x^2 + 3x^3 .

We do need the fact that P ( 1 ) = 6 P(1) = 6 and that the coefficients are non-negative, to narrow down the polynomial more. See Niranjan's solution for the steps to properly do this.

As an explicit example, can you find 2 polynomials with non-negative coefficients that satisfy P ( 1 ) = 7 P(1) = 7 and P ( 5 ) = 651 = 1010 1 5 P(5) = 651 = 10101_5 ?

If however, we had P ( 1 ) < 5 P(1) < 5 , then because (and this is the added justification of the incomplete step) we know that the individual coefficients have to be less than 5, hence the base 5 representation is the unique way to do so, and so we can draw the conclusion immediately.
E.g. If we had 1) coefficients are non-negative, 2) P ( 1 ) = 4 P(1) = 4 , 3) P ( 5 ) = 424 P(5) = 424 , then this approach works (because of my justification.)

Calvin Lin Staff - 2 years, 7 months ago

Log in to reply

@Calvin Lin – Your polynomial in the first half of the claim is wrong but I get your point

Krutarth Patel - 2 years, 7 months ago

Log in to reply

@Krutarth Patel – Thanks, fixed.

Calvin Lin Staff - 2 years, 7 months ago

@Calvin Lin – @Kenny Lau Just in case you didn't get this previously.

Calvin Lin Staff - 2 years, 7 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...