Foundations of Arithmetic: Star problems

Each product has one factor below 50 and the other the same distance above 50.

If that distance is $r$, expand by the distributive property:

The two middle contributions cancel.

Making the factors less equal therefore reduces the product by the square of the distance from 50.

For example, $49\times51$ is 1 below 2500, while $46\times54$ is 16 below 2500.

The four distances are 1, 2, 3, and 4.

Hence the total shortfall from four copies of 2500 is

Four copies of 2500 total 10000, so the desired sum is 9970.

This explains the comparison and avoids four separate multiplication calculations.

First compare neighboring buttons.

Starting with a current value $x$, AD produces $2(x+7)=2x+14$, whereas DA produces $2x+7$.

Thus AD produces 7 more than DA.

Any remaining additions and doublings preserve this advantage, possibly making it larger.

To maximize the final value, move each A left past every D before it.

Each exchange of DA for AD strictly increases the final result.

Eventually both additions come first, so the maximum comes from AADDD.

It is

To minimize the value, make the reverse exchanges until both additions come last.

The order is DDDAA, giving

Every other order contains a neighboring pair that can be exchanged to move toward the appropriate extreme.

This proves that the displayed values are global best values, rather than merely good examples.

For any fixed pairing of the cards, orienting both numbers with their larger digit first gives the largest product for that pairing: both factors are positive, and increasing either one increases their product.

Orienting both with the smaller digit first similarly gives the smallest product.

There are exactly three pairings.

The digit 2 must be paired with 3, 7, or 8; after that, the other pair is forced.

Their largest products are $32\times87=2784$, $72\times83=5976$, and

The greatest is therefore 5986.

For the same three pairings, the smallest products are $23\times78=1794$, $27\times38=1026$, and

The least is 1026.

Each within-pair orientation changes a factor strictly, and the three compared extreme products are different.

Thus no other factor pairs tie these answers, except for swapping the factors.

Each digit appears once in the hundreds place, once in the tens place, and once in the units place.

The total is therefore 111 times the sum of the digits.

Since $1998=111\times18$, the three different digits must sum to 18.

List a candidate set as $a<b<c$.

The smallest digit cannot exceed 5: if it were at least 6, the smallest possible set would be 6, 7, 8, whose sum is already 21.

Now check the possible smallest digits from 1 through 5.

If $a=1$, the other digits sum to 17, forcing 8 and 9.

If $a=2$, they sum to 16, forcing 7 and 9.

If $a=3$, the possibilities are 6 and 9 or 7 and 8.

If $a=4$, they are 5 and 9 or 6 and 8.

If $a=5$, they must be 6 and 7.

These exhaust the increasing pairs within the digit range 1 to 9.

Every listed set works in any order.

There are three choices for the first digit and two for the second, after which the third is fixed: six numbers per set.

Different sets cannot produce the same number, so the count is

There are five numbers.

If the parentheses begin at the first number, they can end at any of the next four numbers; beginning at the second gives three choices, then two, then one.

Thus there are ten allowed spans in total.

Five placements leave the original value $6+20+6=32$ unchanged: parentheses around $2\times3$, around $4\times5$, around $4\times5+6$, around $2\times3+4\times5$, or around the whole expression.

The other five resulting expressions have values $(2\times3+4)\times5+6=56$; $2\times(3+4)\times5+6=76$; $2\times(3+4\times5)+6=52$; $2\times(3+4\times5+6)=58$; and

These five changed placements and five unchanged placements exhaust the ten spans.

The unique greatest value is 76.

The successful placement makes both neighboring multiplications act on the sum $3+4$, so the effect of that addition is amplified on both sides.

Call the original smaller number $s$ and the larger $l$.

Reducing $l$ by 1 first decreases the product by $s$.

Increasing $s$ by 1 next increases the product by the new larger factor, $l-1$.

The net increase is therefore $l-s-1$.

This increase is 12, so the original gap between the numbers is 13.

Remove this extra 13 from their total 83: the remaining 70 consists of two equal copies of the smaller number.

The smaller is 35 and the larger is 48.

Their sum and gap determine them uniquely.

Indeed, $35\times48=1680$ and

For a transfer of 2, reducing the larger factor first loses

Increasing the smaller factor by 2 then gains

The net increase is $92-70=22$.

Equivalently, the new product $37\times46=1702$ is 22 above 1680.

Tracking the two changes avoids guessing the hidden numbers or repeatedly expanding large products.

For an upper bound, temporarily allow any three positive whole-number sums totaling 45.

If two sums are $s$ and $l$, with $l$ at least 2 larger, replacing them by $s+1$ and $l-1$ increases their product by $l-s-1>0$.

Multiplication by the unchanged positive third sum preserves the increase.

Repeating brings the sums as close together as possible: 15, 15, 15.

Thus the product is at most

The three displayed groups each sum to 15, so this bound is achievable.

For the minimum, take a grouping with least product and order its groups by sum.

Suppose a group with smaller or equal sum contains a larger card than a group later in this order.

Swap those two cards.

If their difference is $d>0$, the earlier sum decreases by $d$ and the later increases by $d$.

Their product decreases by $d$ times the old gap between the sums, plus $d\times d$.

The other sum remains positive, so the full product strictly decreases, a contradiction.

Therefore every card in the first group must be smaller than every card in either later group, and every card in the second smaller than every card in the third.

The groups must be {1,2,3}, {4,5,6}, {7,8,9}.

Their sums are 6, 15, 24, giving

Write the digits as $a,b,c,d$ from left to right.

In the units column, $a+d$ ends in 0.

It lies between 2 and 18, so it must be 10, giving a carry of 1.

In the tens column, $b+c+1$ ends in 1 and lies between 3 and 19, so it must be 11.

Thus $b+c=10$, again with carry 1.

The hundreds and thousands columns now also total 11, producing exactly 11110.

The allowable unordered pairs of different nonzero digits summing to 10 are {1,9}, {2,8}, {3,7}, and {4,6}.

The pair {5,5} is excluded because the digits must differ.

Choose the outside pair in four ways and the inside pair in three remaining ways.

Each pair has two orientations, so there are $4\times3\times2\times2=48$ numbers.

The pairs are disjoint, and each number determines these choices uniquely.

To minimize, start with digit 1; the last digit is then 9.

The smallest available second digit is 2, forcing the third to be 8.

This gives 1289.

Similarly, the maximum begins 9, ends 1, and uses 8 then 2 inside, giving 9821.

Consider two weights, one smaller and one larger, and two cards, also one smaller and one larger.

Matching the larger card to the larger weight gives a score greater than the crossed matching by the difference between the weights multiplied by the difference between the cards.

This is positive.

For instance, weights 3 and 8 with cards 4 and 9 give $3\times4+8\times9=84$, compared with $3\times9+8\times4=59$; the gain is

The distributive property gives the same rule for any pair.

At a maximum, there can be no pair of cards out of increasing order, because exchanging that pair would increase the score.

Thus the cards must increase with the weights.

The score is

At a minimum, there can be no pair of cards in increasing order: putting the larger card at the smaller weight would reduce the score.

The cards must therefore decrease as the weights increase.

This gives

All weights and cards differ, so each optimal assignment is unique.