## Use the given statements in the right way to limit possible values of the digits quickly

### The problem: 5th Hard number system question for CAT

If 495 is subtracted from a 3-digit number with its digits reversed, result will be the original number. Sum of the digits of the number is 12 and sum of all pairwise products of its digits is 41. Find the number.

**Hint:** For quick solution of the 5th hard number system question for CAT, limit the possible values of the digits by identifying breakthrough conditions and enumeration using the given conditions.

### Solution to the 5th hard number system question for CAT: Step by step limiting constraints on the digit values by variable value linking technique

Let the required number be,

$100a+10b+c$, that is, $abc$.

Identify the first statement to have highest potential for directly limiting values of the digits by the given relation and the condition that all three digits are less than 10.

From the first statement,

$100a+10b+c=100c+10b+a-495$, number with digits reversed is $(100c+10b+a)$.

Simplifying,

Or, $495=99(c-a)$,

Or, $c-a=5$.

This important relation restricts the values of the two digits $a$ and $c$, both less than 10, to,

$a=1$, $c=6$,

$a=2$, $c=7$,

$a=3$, $c=8$, and,

$a=4$, $c=9$.

What we have done is enumeration to limit the possible values of the two digits.

Now in the second step we will limit the values of all three digits by the second statement.

By the second statement, the sum of the three digits is 12.

So,

$a+b+c=12$

Substitute $a+5$ for $c$ eliminating $c$ for limiting the values of the remaining digit $b$,

$a+b+a+5=12$,

Or, $2a+5=7$

Applying the possible values of $a$ in the already enumerated possible value pairs on this equation, the values of all three digits are limited to,

$a=1$, $b=5$, $c=6$..........Combination 1

$a=2$, $b=3$, $c=7$..........Combination 2

$a=3$, $b=1$, $c=8$..........Combination 3.

$a=4$ is not possible as value of $b$ will then be out of valid range.

This is as fast as we can limit the values.

Now we will use the third relation,

$ab+bc+ca=41$.

Calculating the values of $(ab+bc+ca)$ for the three digits for each combination,

**Combination 1:** $a=1$, $b=5$, $c=6$: $ab+bc+ca=5+30+6=41$

**Combination 2:** $a=2$, $b=3$, $c=7$: $ab+bc+ca=6+21+14=41$

**Combination 3:** $a=3$, $b=1$, $c=8$: $ab+bc+ca=3+8+24=35 \neq 41$

Two possible numbers corresponding to Combinations satisfy all the given conditions: 156 and 217.

**Answer:** 156 and 217.

**Note:** Solution speed would depend on how confidently you select the first relation as the starting point and proceed from there.

The **only problem solving technique** used is the **Variable value linking technique.**

### End note

In solving reasoning puzzles, we had to use **an abstract form** of the technique, the **linking bonded member structures.**

Here, the digits are the members of the structure of each Combination in which they are bonded or fixed. The bonded structures or Combinations are expanded step by step by linking with the other *similar structures bonded in the equations.*

In simple terminology, we have named this process as, **Variable value linking technique** which is good enough in this case.

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