UKMT Special (Problem $22$ )

Each of $A$ and $B$ is a $4$ -digit palindromic integer.

$C$ is a $3$ -digit palindromic integer and

$A - B = C$

What are the possible values of $C$ ?

[UKMT Hamilton Olympiad $2019$ , H $4$ ]

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Math	Appears as
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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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Comments

You have until next Saturday, $3:00$ pm!

Yajat Shamji - 6 months, 3 weeks ago

Rewrite as

   a b b a
   c d d c
   _______-
     e f e

Step 1) Notice that the difference of 2 four digit number is a three digit number, this implies that (look at thousand column) a - c = 0 or a - c - 1 = 0 (in the case of hundred column borrowing from thousands column).

   If a - c = 0, then (look at ones column) e = 0 
   and implies that the subtraction result is a two digit number which is not true. 
   Therefore, a - c - 1 = 0   ->   a = c + 1.

Step 2) Notice that (look at ones column) a - c = e or 10 + a - c = e (in the case of hundred column borrowing from tens column).

   Substitute a from step 1.
   If 10 + a - c = e, then 10 + a - c = e     ->     10 + c + 1 - c = e     ->     e = 11 
   However, c is a one digit, so this assumption must be false.
   Therefore, a - c = e     ->     c + 1 - c = e     ->     e = 1

   a b b a
   c d d c
   _______-
     1 f 1

Step 3) Notice that a = c + 1, however, (look at thousands column) the result is three digit, this implies that b < d and borrow from the thousands column , 10 + b - d = 1 or 10 + b - d - 1 = 1

   If 10 + b - d = 1, this implies that the tens column doesn't borrow, however, b < d, 
   so this assumption must be false.
   Therefore, 10 + b - d - 1 = 1     ->     8 + b = d       ->     b = 0, d = 8  or  b = 1, d = 9
   Otherwise, d is a two digit number

Step 4) Notice that, since a = c + 1 the ones column doesn't borrow from the tens column and b < d, therefore 10 + b - d = f (in the case of hundred column borrowing from tens column).

   Substitute b and d from step 3.
   Case 1 : b = 0, d = 8 
   10 + 0 - 8 = f     ->     f = 2
   Case 2 : b = 1, d = 9
   10 + 1 - 9 = f     ->     f = 2
   In either case, f = 2

   a b b a
   c d d c
   _______-
     1 2 1

Therefore, C = 121

Quest Keeper - 6 months, 3 weeks ago

That's wrong. Try again!

Yajat Shamji - 6 months, 3 weeks ago

I see, I forgot the carrier. BTW, do you do integral?

Quest Keeper - 6 months, 3 weeks ago

@Quest Keeper – No.

Yajat Shamji - 6 months, 3 weeks ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

UKMT Special (Problem 222222)

Comments

UKMT Special (Problem $22$ )