word problems

[mmarti25]

1. Take a two-digit integer. Multiply it by the integer you get when you interchange its digits, then subtract 10 times the sum of the squares of the original digits. The result is always divisible by 101. Why?

2. Suppose r is a solution of a*x^2 + b*x + c = 0, where a*c does not = 0. Show that 1/r is a solution of c*x^2 + b*x + a = 0.

Please help!

 

[tkhunny]

1. Take a two-digit integer. Multiply it by the integer you get when you interchange its digits, then subtract 10 times the sum of the squares of the original digits. The result is always divisible by 101. Why?

First, test it and see if it EVER works. Let's try 10.

(10)*(1) - 1 = 9 -- Well, that didn't work. Maybe it means everything is non-zero. Maybe 12.

(12)*(21) - 5 = 247 -- That didn't work, either.

This leaves us with only a couple of choices. The problem has not been presented correctly or in its entirety, or we were supposed to waste a lot of time on algebra for something that simply isn't valid.

 

[Denis]

Sure it works, TK:

(10)*(1) - 1 = 9 -- Well, that didn't work.

10 * 1 = 10 - 10 * 1 = 0 : 0 / 101 = 0

(12)*(21) - 5 = 247 -- That didn't work, either.

12 * 21 = 252 - 10*5 = 202: 202 / 101 = 2

Better TaKe 5 :roll:

 

[happy]

You guys are a riot!!!!!! :lol: :lol: :lol:

 

[tkhunny]

Maybe I'd better take "10" on this one.

 

[ting]

Let the digits in the number be, n, and m, then

(10n +m)(10m + n) - 10(n^2 + m^2)


= 100mn + 10m^2 + 10n^2 +mn - 10n^2 + 10m^2

= 101 mn

 

[soroban]

Hello, mmarti25!

1. Take a two-digit integer.
(1) Multiply it by the integer you get when you interchange its digits.
(2) Then subtract 10 times the sum of the squares of the original digits.
The result is always divisible by 101. Why?

Let the original two-digit number be: .10t + u.

(1) Interchange the digits and we have: .10u + t.
. . . Then: .(10t + u)(10u + t) .= .10t² + 101tu + 10u²

(2) The sum of the squares of the digits is: .t² + u²
. . . Then: .(10t² + 101tu + 10u²) - 10(t² + y²) .= .101tu

Since t and u are digits (whole numbers), then 101tu is divisible by 101.

[Edit: Too fast for me, ting!]

2. Suppose r is a solution of ax² + bx + c = 0, where ac ≠ 0.
Show that 1/r is a solution of: cx² + bx + a = 0.

Since r is a solution of ax² + bx + c = 0, then: .ar² + br + c .= .0
. . . . . . . . . . . . . . .b . . .c
Divide by r²: . a + -- + --- . = . 0
. . . . . . . . . . . . . . .r . . .r²

We have: . c(1/r)² + b(1/r) + a .= .0

Therefore, x = 1/r is a solution of: .cx² + bx + a .= .0