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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 $a.b - c.d$ ?

User Vibeke
by
6.1k points

2 Answers

4 votes

Answer:

18.3

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.

User Mathieu Castets
by
5.9k points
5 votes

Answer:

70

Explanation:

  • for the value of 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 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