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Decimals: Star problems

Ten optional challenges to stretch your reasoning. Work on paper, use hints when you need them, and check the answer or full solution when you are ready. You can skip these problems and continue the course.

1 of 3 stars: Stretch
2 of 3 stars: Challenge
3 of 3 stars: Deep challenge

Stars indicate difficulty within this set.

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Problem 1 of 10

Problem 1 One hundredth closer

Difficulty: 1 of 3 stars, Stretch

Two rectangular tiles have side lengths 0.48 m and 0.52 m, and 0.49 m and 0.51 m, respectively.

Which tile has greater area, and by exactly how much? Justify your answer without calculating both complete products.
1. Hint 1
  Compare the two products with the shared product $0.48\times0.51$ .
2. Hint 2
  The extra strips have the same width, 0.01 m, but different lengths.
Answer

The $0.49$ m by $0.51$ m tile is larger by $0.0003$ square meters.

Full solution

Both products contain the same part, $0.48\times0.51$ .
The first tile adds a strip with area $0.48\times0.01$ , while the second adds a strip with area $0.51\times0.01$ .
The second extra strip is larger.

Subtract only these extra pieces.
The difference is $(0.51-0.48)\times0.01=0.03\times0.01=0.0003$ square meters.
The common area cancels, so we never need either full product.

The tiles have the same perimeter because their side lengths have the same sum.
Moving the lengths closer together increases the area in this comparison.
The reason is precise: the gain from extending the shorter side is larger than the loss from shortening the longer side.

Answer

The $0.49$ m by $0.51$ m tile is larger by $0.0003$ square meters.

Key idea

Compare complicated quantities by first removing a shared part.
Problem 1: Mark complete
Problem 2 The third hiker's share

Difficulty: 1 of 3 stars, Stretch

Anika brings 2.40 kg of trail mix and Ben brings 1.35 kg of the same trail mix. Cara brings none. The three hikers divide all the trail mix equally and eat their shares.

Cara pays 7.50 dollars for the trail mix she receives. Each kilogram has the same price. How much of this payment should Anika receive, and how much should Ben receive? Explain why dividing the payment in proportion to their original contributions would be incorrect.
1. Hint 1
  First work out how much each hiker eats.
2. Hint 2
  For each supplier, subtract the amount that supplier eats before finding how much is sold to Cara.
Answer

Anika receives 6.90 dollars; Ben receives 0.60 dollars.

Full solution

The total is $2.40+1.35=3.75$ kg.
Each hiker receives $3.75\div3=1.25$ kg.
Anika eats 1.25 kg of her own contribution, leaving $2.40-1.25=1.15$ kg for Cara.
Ben eats 1.25 kg of his own contribution, leaving only $1.35-1.25=0.10$ kg for Cara.

Cara therefore buys 1.25 kg altogether.
Her payment of 7.50 dollars means the price is $7.50\div1.25=6$ dollars per kilogram.
Anika receives $1.15\times6=6.90$ dollars; Ben receives $0.10\times6=0.60$ dollars.
These payments add to 7.50 dollars.

The original contributions include food that Anika and Ben themselves consume.
Cara pays only for the portions transferred to her.
Tracking that transfer, rather than the initial amounts, determines the fair split.

Answer

Anika receives 6.90 dollars; Ben receives 0.60 dollars.

Key idea

Identify which quantity is actually being shared or paid for.
Problem 2: Mark complete
Problem 3 A forbidden digit

Difficulty: 1 of 3 stars, Stretch

Write every positive decimal from 0.01 through 0.99 with exactly two digits after the decimal point. Cross out every number whose written form contains the digit 5.

Find the sum of all the remaining numbers without adding them individually. Explain how you count each digit contribution.
1. Hint 1
  Include 0.00 temporarily; adding zero does not change the sum.
2. Hint 2
  In each decimal position, each allowed digit appears equally often.
Answer

39.60.

Full solution

Temporarily include 0.00.
Each decimal position can contain any of the nine digits 0, 1, 2, 3, 4, 6, 7, 8, 9.
Their sum is $45-5=40$ .
Choosing a digit for one position leaves nine choices for the other position.

Consequently, each allowed digit occurs nine times in the tenths position.
The total contribution from that position is
$9\times40\times0.1=36.$
Each allowed digit also occurs nine times in the hundredths position, contributing
$9\times40\times0.01=3.6.$

The required total is
$36+3.6=39.6.$
We have counted 81 decimal strings, one of which is 0.00.
Removing that string leaves the 80 positive numbers requested and does not change the sum.
This method separates place value from the particular order of the numbers.

Answer

39.60.

Key idea

Sum contributions by place value when a long list has symmetry.
Problem 3: Mark complete
Problem 4 Can rounded totals disagree?

Difficulty: 2 of 3 stars, Challenge

Two positive measurements are exact multiples of 0.001. When each is rounded to the nearest tenth, the displayed measurements are 2.4 and 3.7.

Add the two exact measurements first and then round their sum to the nearest tenth. Find every possible displayed total. Give an example for each and prove that your list is complete.

Use round-half-up throughout: an exact halfway value rounds to the larger tenth.
1. Hint 1
  Find the smallest and largest three-decimal measurements that can display as 2.4.
2. Hint 2
  Bound the exact sum before deciding which rounding intervals it can reach.
Answer

The possible displayed totals are 6.0, 6.1, and 6.2.

Full solution

The first exact measurement can be any thousandth from 2.350 through 2.449.
The second can be any thousandth from 3.650 through 3.749.
The lower endpoints round up to the stated displays; the next thousandths above the upper endpoints would round to the next tenths.

The exact sum therefore lies from $2.350+3.650=6.000$ through
$2.449+3.749=6.198.$
Rounding a number in this interval to the nearest tenth can give only 6.0, 6.1, or 6.2.
In particular, a display of 5.9 requires a sum below 5.950, while 6.3 requires at least 6.250.

All three candidates occur: 2.350 and 3.650 give 6.000, displaying 6.0; 2.400 and 3.700 give 6.100, displaying 6.1; 2.449 and 3.749 give 6.198, displaying 6.2.
These examples establish possibility, and the bounds establish completeness.

Answer

The possible displayed totals are 6.0, 6.1, and 6.2.

Key idea

A rounded value describes an interval, not a single exact measurement.
Problem 4: Mark complete
Problem 5 The double-rounding trap

Difficulty: 2 of 3 stars, Challenge

Consider the 1,000 numbers 0.000, 0.001, 0.002, ..., 0.999.

Method A rounds a number directly to the nearest tenth. Method B first rounds it to the nearest hundredth, then rounds that result to the nearest tenth. Both methods use round-half-up.

Find every number for which the methods disagree. Describe the complete set compactly, count its members, and state which method gives the larger result.
1. Hint 1
  Look just below the halfway point 0.050.
2. Hint 2
  Repeat the same argument in each interval of length 0.1.
Answer

For each digit $d$ from 0 through 9, the numbers $\frac{d}{10}+0.045$ , $\frac{d}{10}+0.046$ , ..., $\frac{d}{10}+0.049$ disagree. There are 50; Method B is larger by 0.1.

Full solution

In the first tenth, direct rounding changes from 0.0 to 0.1 at 0.050.
However, the five numbers 0.045 through 0.049 first round to 0.05.
The second rounding then sends them to 0.1, although direct rounding sends them to 0.0.

No other thousandth from 0.000 through 0.099 causes disagreement.
Values at most 0.044 first round to at most 0.04 and still finish at 0.0.
Values at least 0.050 round to 0.1 directly and also finish at 0.1 after the two stages.

The same argument applies after adding 0.1, 0.2, and so on through 0.9.
Thus the full set is 0.045-0.049, 0.145-0.149, ..., 0.945-0.949, with steps of 0.001 in each block.
There are ten blocks of five numbers, or 50.
In every case Method B raises the final answer by one tenth; it never lowers it.

Answer

For each digit $d$ from 0 through 9, the numbers $\frac{d}{10}+0.045$ , $\frac{d}{10}+0.046$ , ..., $\frac{d}{10}+0.049$ disagree. There are 50; Method B is larger by 0.1.

Key idea

Rounding can move a value onto a later rounding boundary.
Problem 5: Mark complete
Problem 6 Reverse the repeating block

Difficulty: 2 of 3 stars, Challenge

Two distinct digits $a$ and $b$ are chosen from 1 through 9. Let $x=0.ababab\ldots$ and $y=0.bababa\ldots$ , where each two-digit block repeats forever.

The numbers satisfy $x+y=1$ , and $x$ exceeds $y$ by $3/11$ . Find the digits and justify that there is only one answer.
1. Hint 1
  A decimal with a two-digit repeating block can be written as that two-digit integer divided by 99.
2. Hint 2
  Use the sum to find $a+b$ and the difference to find $a-b$ .
Answer

$a=6$ and $b=3$ ; $x=7/11$ and $y=4/11$ .

Full solution

Moving the decimal point two places right shifts one complete block before the point.
Subtracting the original decimal cancels the repeating tails.
Thus $99x=10a+b$ , so
$x=(10a+b)/99.$
Similarly,
$y=(10b+a)/99.$

Adding gives
$x+y=11(a+b)/99=(a+b)/9.$
Since the sum is 1, the digits add to 9.
Subtracting gives
$x-y=9(a-b)/99=(a-b)/11.$
Since this difference is $3/11$ , the first digit exceeds the second by 3.

Two numbers with sum 9 and difference 3 must be 6 and 3: remove the extra 3 from the sum, then divide the remaining 6 equally.
Checking, $63/99=7/11$ and $36/99=4/11$ , whose sum is 1 and difference is $3/11$ .
The sum and difference fix both digits, proving uniqueness.

Answer

$a=6$ and $b=3$ ; $x=7/11$ and $y=4/11$ .

Key idea

Convert a repeating block to a fraction before comparing it.
Problem 6: Mark complete
Problem 7 A one-dollar rounding gap

Difficulty: 2 of 3 stars, Challenge

A supplier sells small components at an exact price of 0.074 dollars each. An order contains a positive whole number of components.

Invoice A rounds the price of one component to the nearest cent and then multiplies by the number ordered. Invoice B multiplies using the exact price and rounds only the final total to the nearest cent. All rounding uses round-half-up.

What is the smallest order for which Invoice B is at least 1.00 dollar greater than Invoice A? Prove that every smaller order fails.
1. Hint 1
  Work in cents: the exact price is 7.4 cents, while Invoice A charges 7 cents per component.
2. Hint 2
  How large must the unrounded excess be to round up to 100 cents?
Answer

249 components. Invoice A is 17.43 dollars and Invoice B is 18.43 dollars.

Full solution

Let the number ordered be $n$ .
In cents, Invoice A charges the whole number $7n$ .
Invoice B rounds $7.4n=7n+0.4n$ to a whole number of cents.
Since $7n$ is already a whole number, the gap between the invoices is simply $0.4n$ rounded to a whole number of cents.

That rounded gap first reaches 100 cents when its unrounded value reaches 99.5 cents.
For 248 components the unrounded gap is $0.4\times248=99.2$ cents, which rounds to 99 cents.
Every smaller order has an even smaller unrounded gap, so none can work.

For 249 components, the gap before rounding is 99.6 cents, which rounds to 100 cents.
Directly, Invoice A is $249\times0.07=17.43$ dollars.
Invoice B begins at $249\times0.074=18.426$ dollars and rounds to 18.43 dollars.
The difference is exactly 1.00 dollar, proving both feasibility and minimality.

Answer

249 components. Invoice A is 17.43 dollars and Invoice B is 18.43 dollars.

Key idea

When rounding an error, compare it with the half-unit threshold.
Problem 7: Mark complete
Problem 8 Four rotating decimals

Difficulty: 3 of 3 stars, Deep challenge

Choose four distinct digits $a,b,c,d$ from 0 through 9; zero is allowed. Form four infinite repeating decimals:

$0.abcdabcd\ldots$ , $0.bcdabcda\ldots$ , $0.cdabcdab\ldots$ , and $0.dabcdabc\ldots$ . Their sum is exactly 2.

Find the greatest possible value of the first decimal. Prove that your choice satisfies the sum condition and that no larger choice works.
1. Hint 1
  Across the four decimals, what is the combined contribution in any one decimal position?
2. Hint 2
  After finding the required digit sum, maximize the first digit, then the second.
Answer

$0.98109810\ldots$ , with $(a,b,c,d)=(9,8,1,0)$ .

Full solution

Let the sum of the four digits be $D$ .
Across all four decimals, each position contains all four digits once.
An exact fractional way to combine them is to place the four repeating four-digit blocks over 9999.
Their numerators sum to $1111D$ , because each digit occupies each place once.
The total is therefore
$1111D/9999=D/9.$
The required total of 2 means $D=18$ .

To maximize the first decimal, choose its first digit as large as possible.
The digit 9 is feasible: the distinct remaining digits 8, 1, and 0 sum to 9.
With 9 fixed first, the largest distinct second digit is 8, and that choice is feasible too.

The final two digits must now be distinct and sum to 1, so they are 1 and 0.
Putting 1 first gives the larger decimal.
Thus the maximum is $0.98109810\ldots$ .
Its digits sum to 18, so the four rotations sum to 2.
Any competing decimal with a smaller first differing digit is smaller, proving maximality.

Answer

$0.98109810\ldots$ , with $(a,b,c,d)=(9,8,1,0)$ .

Key idea

Turn a positional symmetry into a constraint, then optimize from left to right.
Problem 8: Mark complete
Problem 9 Use every digit once

Difficulty: 3 of 3 stars, Deep challenge

Use the digits 0 through 9 exactly once to form five decimal numbers of the form $0.ab$ . Each number has exactly two digits after the point; examples such as 0.04 and 0.70 are allowed. The fixed zero before each decimal point does not use a digit from your supply.

(a) Determine every possible sum of the five numbers. Give a compact description and prove that no possible sum is missing.

(b) If the sum rounds to 2.5 to the nearest tenth, using round-half-up, what must its exact value be? Exhibit an arrangement.
1. Hint 1
  Let $T$ be the sum of the five digits placed in the tenths positions. The other five digits sum to $45-T$ .
2. Hint 2
  Start with tenths digits 0,1,2,3,4. Can you increase their sum one unit at a time until you reach 5,6,7,8,9?
Answer

(a) The 26 totals $1.35+0.09k$ , where $k=0,1,\ldots,25$ . (b) $2.52$ , for example $0.01+0.23+0.54+0.76+0.98$ .

Full solution

Let $T$ be the sum of the tenths digits.
All ten digits sum to 45, so the hundredths digits sum to $45-T$ .
The total is
$0.1T+0.01(45-T)=0.45+0.09T.$
The smallest possible $T$ is $0+1+2+3+4=10$ and the largest is
$5+6+7+8+9=35.$

Every whole-number value from 10 through 35 is achievable.
Begin with tenths digits 0,1,2,3,4.
Increase the last digit successively to 9, then the next-to-last digit successively to 8, then the middle digit to 7, the second digit to 6, and the first digit to 5.
At every step the digits stay distinct and their sum increases by exactly 1.
The unused digits can always fill the hundredths positions.
This proves that all 26 stated totals occur.

To round to 2.5, the exact sum must be at least 2.45 and below 2.55.
The nearby possible totals are 2.43, 2.52, and 2.61, so only 2.52 works.
The arrangement $0.01+0.23+0.54+0.76+0.98=2.52$ uses every supplied digit once.

Answer

(a) The 26 totals $1.35+0.09k$ , where $k=0,1,\ldots,25$ . (b) $2.52$ , for example $0.01+0.23+0.54+0.76+0.98$ .

Key idea

A complete answer needs both a restriction and a construction for every permitted case.
Problem 9: Mark complete
Problem 10 A thousand tiny roundings

Difficulty: 3 of 3 stars, Deep challenge

A data file contains each of the 1,000 numbers 0.000, 0.001, 0.002, ..., 0.999 exactly once.

(a) Find the exact sum of the file.

(b) Round every entry directly to the nearest tenth and then add. Find the new total.

(c) Instead round every original entry to the nearest hundredth, then round that result to the nearest tenth, and then add. Find this total.

Use round-half-up throughout. Explain all three totals without adding 1,000 entries individually, and explain why the rounding errors do not cancel.
1. Hint 1
  For the original sum, pair the smallest entry with the largest.
2. Hint 2
  For direct rounding, count how many entries display as 0.0, 0.1, ..., 1.0. For double rounding, inspect the five thousandths just below each halfway boundary.
Answer

(a) 499.5. (b) 500.0. (c) 505.0.

Full solution

Pair 0.000 with 0.999, 0.001 with 0.998, and continue inward.
There are 500 pairs, each totaling 0.999, so the exact sum is
$500\times0.999=499.5.$

Under direct rounding, the 50 entries from 0.000 through 0.049 display as 0.0.
Each display from 0.1 through 0.9 receives 100 entries; for example, 0.050 through 0.149 display as 0.1.
The final 50 entries, 0.950 through 0.999, display as 1.0.
The total is
$100(0.1+0.2+\ldots+0.9)+50=100\times4.5+50=500.$

Double rounding changes precisely the five thousandths immediately below each boundary 0.050, 0.150, ..., 0.950.
They first round onto the boundary and then round upward.
There are 50 such entries, and each increases by 0.1 compared with direct rounding.
The new total is therefore
$500+50\times0.1=505.$

The direct-rounding errors have a net upward total of 0.5, because halfway ties consistently go upward.
Double rounding adds another 5.0 by moving extra entries onto those upward-rounding boundaries.
A small error per entry can accumulate systematically.

Answer

(a) 499.5. (b) 500.0. (c) 505.0.

Key idea

Study error patterns across the whole data set; small errors need not cancel.
Problem 10: Mark complete