Using 1-9 digits (2)

Number Theory Level 5

If you use all the digits from 1 to 9 to make up three, 3-digit numbers such that their product will be maximum and then remake other three, 3-digit numbers such that their product will be minimum. What will be the difference between the maximum product and the minimum product?

The answer is 597726822.

1 solution

Karim Fawaz
Dec 28, 2016

To find the minimum product:

The first digit of each of the three-digit numbers must be the smallest possible one; hence we may assume that the three numbers have the form: 1Aa, 2Bb, 3Cc where the overbar indicates a succession of digits; for example,

$2Bb = 2 X 10^{2} + B X 10 + b.$

We shall show that necessarily the following 3 conditions must be met:

A < B < C
a < b < c
each of the digits a, b, and c is greater than any of the digits A, B, and C.

Proof of condition 1:

Assume A > B. Then Aa > Bb, and so

1Aa X 2Bb - 2Aa X 1Bb = (100 + Aa) (200 + Bb) - (200 + Aa) (100 + Bb) = 100 (Aa - Bb) > 0,

which would mean that: 1Bb X 2Aa X 3Cc < 1Aa X 2Bb X 3Cc,

whence the product on the right would not be the least, as required. The assumption B > C will produce a similar contradiction.

From the above we conclude that A < B < C. (1)

Proof of condition 2:

Assume a > b. Then

1Aa X 2Bb - 1Ab X 2Ba = (10.1A+ a) (10·2B + b) - (10.1A + b) (10·2B + a) = (10·2B - 10· 1A) (a - b) > 0,

which would mean that: 1Ab X 2Ba X 3Cc < 1Aa X 2Bb X 3Cc

which again is a contradiction. The result b < c is similarly shown.

From the above we conclude that a < b < c. (2)

Proof of condition 3:

Assume C > a, or C = a + x, where x > 0.

According to the first demonstration C is the largest of the digits A, B, C; and according to the second demonstration a is the smallest of the digits a, b, c. In this case we would have 1Aa X 3Cc - 1AC X 3ac = 1Aa (3ac + 10x) - (1Aa + x) (3ac) = x (10· 1Aa - 3ac) > 0 ,

which yields the contradiction 1AC X 2Bb X 3ac < 1Aa X 2Bb X 3Cc

From the above we conclude that (a, b, c) > (A, B, C). (3)

It follows from (1), (2), and (3) that: A < B < C < a < b < c

This means that: A = 4, B = 5, C = 6, a = 7, b = 8 and c = 9. whence the minimum product sought is composed as follows: 147 X 258 X 369 = 13,994,694.

To find the maximum product:

It is clear that the largest digits must be the initial digits of each number; hence we may write the product in the form 9Aa X 8Bb X 7Cc

Employing techniques analogous to those used in the minimum product, we readily prove that: A < B < C; a < b < c; each of the digits a, b, c is smaller than any of the digits A, B, C. That is: a<b<c<A<B<C.

That means a = 1, b = 2, c = 3, A = 4, B = 5 and C = 6.

And so the maximum product sought is composed as follows: 941 X 852 X 763 = 611,721,516.

The difference between the 2 products is then: 611,721,516 - 13,994,694 = 597,726,822.

$Answer = \boxed{597726822}$

The solution can be simplified by highlighting the important aspects of minimizing the product of 2 terms, namely

Each number should have digits in increasing order (from left to right)
Let $A = a_1 + a_2, B = b_1 + b_2$ . $(a_1 + a_2) ( b_1 + b_2) < ( a_1 + b_2 ) ( b_1 + a_2 ) \Leftrightarrow (a_1 - b_1 ) ( a_2 - b_2 ) > 0$ . This implies that if we let $a_2, b_2$ be the unit/ten/hundreds digit that we want to swap out, then $(a_1, b_1)$ and $(a_2, b_2 )$ have to be similarly ordered.

With this crux, it is clear that the numbers we want are 147, 258, 369 for the minimum. (And of course, your 3 conditions also follow)

In the case of maximizing, the conditions then become

Each number should have digits in decreasing order (from left to right)
If we let $a_2, b_2$ be the unit/ten/hundreds digit that we want to swap out, then $(a_1, b_1)$ and $(a_2, b_2 )$ have to be oppositely ordered.

With this, it is clear that the numbers we want are 941, 852, 763.

Calvin Lin Staff - 4 years, 5 months ago

That is right. For minimum product, the digits have to be in ascending order to have the lowest possible numbers using each 3 digits and for maximum product, the digits have to be in decreasing order to have the highest possible numbers. This condition is of course necessary but not sufficient. Because we can make other 3 numbers which are also in ascending order but they don't give the minimum product. For example, we can have the numbers 148, 259 and 367 which are in ascending order but they don't give the minimum product since 148 X 259 X 367 = 14,067,844 which is bigger than the minimum product of 13,994,694.

Karim Fawaz - 4 years, 5 months ago

But your example doesn't satisfy the second condition. Namely let $A = 140 + 8, B = 360 + 7$ , and we conclude that $(140 + 8 ) ( 360 + 7 ) > ( 140 + 7 ) ( 360 + 8 )$ .

(For an elaboration of my 2nd condition) If we know that the hundreds digit are ordered, then so must the "hundreds + tens" part (which forms $a_1$ ), and hence the digits part (which forms $a_2$ ) must be similarly ordered.

Calvin Lin Staff - 4 years, 5 months ago

Using 1-9 digits (2)

The answer is 597726822.

1 solution

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