If $a$, $b$, $c$, and $d$ are replaced by four distinct digits from $1$ to $9$, inclusive, then what's the largest possible value of the difference $a. b - c. d$ ?

Step-by-step explanation:

Correct answer

Given that a, b, c, and d are different numbers (from 1 to 9), we are to find the largest possible value of a.b + c.d.

This problem can be answered by trial and error and with some logic. Clearly, a to d cannot be at the lower end of the values (1 to 5). Since the digits must be different, it can be a combination of digits from 6 to 9.

It is rather tempting to say that the 9.8 + 7.6 would yield the highest possible sum (=17.4) but this is incorrect. Since it is the sum of two numbers with a decimal, we have to maximize on the one-digit first before maximizing the tenths digit. Therefore: the combination of numbers must then be:

9.7 + 8.6 which results to 18.3

9.6 + 8.7 yields 18.3 as well.

Dollar replacement

For the value of the difference to be largest, the minuend should be maximum(most possibly) and the subtrahend should be minimum

[in A-B=X, A is minuend and B is subtrahend ]

So, $a.b should be the maximum. as there is a condition that 4 digits should be distinct, the product will be maximum if we choose 2 maximum valued numbers from the given numbers. so, one of them should be 9 and the other should be 8.

Therefore, $a.b=9*8=72

As mentioned above, c.d$ should be minimum. this will be possible only when we choose  2 minimum valued numbers from the given numbers. so, one of them should be 1 and the other should be 2.

Therefore, c.d$ = 1*2 = 2

Hence, the difference = 72-2 = 70

Thus,  the largest possible value of the difference $a.b - c.d$ = 70