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Exponents and Roots: 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 A square hiding in a product

Difficulty: 1 of 3 stars, Stretch

Without calculating the product $48\times50$ first, find $\sqrt{48\times50+1}$ .

Then explain why $\sqrt{n(n+2)+1}=n+1$ for every nonnegative whole number $n$ . Your explanation should show the structure of the square.
1. Hint 1
  Compare the two factors with the number halfway between them.
2. Hint 2
  Expand $(n+1)(n+1)$ using the distributive property.
Answer

The value is 49. In general, the square root is $n+1$ .

Full solution

The factors 48 and 50 lie one below and one above 49.
Multiplying gives
$(49-1)(49+1)=49\times49+49-49-1=49^2-1.$
Adding 1 therefore restores the perfect square $49^2$ , whose nonnegative square root is 49.

For the general statement, distribute the multiplication:
$n(n+2)+1=n^2+2n+1.$
Meanwhile,
$(n+1)(n+1)=n^2+n+n+1=n^2+2n+1.$
The expression under the root is exactly $(n+1)^2$ .

A square root means the nonnegative root.
Because $n$ is nonnegative, $n+1$ is positive, so the required root is indeed $n+1$ .
The case $n=0$ works as well: the expression is
$\sqrt{1}=1.$
The key was to recognize the complete square before doing large arithmetic.

Answer

The value is 49. In general, the square root is $n+1$ .

Key idea

Numbers equally spaced around a center often hide a square.
Problem 1: Mark complete
Problem 2 Three power cards

Difficulty: 1 of 3 stars, Stretch

Seven cards are labeled $2^1,2^2,2^3,2^4,2^5,2^6,2^7$ . Choose three different cards whose product is $2^{12}$ .

Find every possible selection. Among them, which has the smallest sum of card values? Justify that your list is complete.
1. Hint 1
  Multiplying powers of 2 adds their exponents.
2. Hint 2
  Order the three exponents from smallest to largest and consider the smallest one first.
Answer

The exponent sets are $(1,4,7)$ , $(1,5,6)$ , $(2,3,7)$ , $(2,4,6)$ , and $(3,4,5)$ . The minimum sum is $8+16+32=56$ .

Full solution

Call the three increasing exponents $a<b<c$ .
Their product is $2^{a+b+c}$ , so their sum must be 12.
The smallest exponent cannot be 4 or greater, because even $4+5+6=15$ is too large.

If $a=1$ , the other two exponents sum to 11.
Within 2 through 7, the increasing pairs are $(4,7)$ and $(5,6)$ .
If $a=2$ , the other two sum to 10 and must exceed 2; the pairs are $(3,7)$ and $(4,6)$ .
If $a=3$ , the remaining sum is 9 and both exponents exceed 3, leaving only $(4,5)$ .
This exhausts all possibilities.

The corresponding sums of card values are $2+16+128=146$ , $2+32+64=98$ , $4+8+128=140$ , $4+16+64=84$ , and
$8+16+32=56.$
The last selection has the smallest sum.
Notice that a fixed product does not fix the sum: how evenly the factors are distributed matters.

Answer

The exponent sets are $(1,4,7)$ , $(1,5,6)$ , $(2,3,7)$ , $(2,4,6)$ , and $(3,4,5)$ . The minimum sum is $8+16+32=56$ .

Key idea

Replace a product of powers by a sum of exponents, then organize the cases.
Problem 2: Mark complete
Problem 3 Half of what remains

Difficulty: 1 of 3 stars, Stretch

A square starts completely white. At each stage, exactly half of the currently white area is shaded. Previously shaded area stays shaded.

What is the first stage at which more than 0.999 of the square is shaded? Give an exact argument without adding all the newly shaded fractions or using decimal approximations to large powers.
1. Hint 1
  Track the white area rather than the shaded area.
2. Hint 2
  Compare the nearby powers of 2 with 1,000.
Answer

Stage 10.

Full solution

After one stage, half the square remains white.
After two, half of that half remains, or $1/4$ .
Continuing, the white fraction after $n$ stages is
$1/2^n=2^{-n}.$
This follows because each stage multiplies the remaining fraction by $1/2$ .

More than 0.999 shaded means less than $1-0.999=0.001=1/1000$ remains white.
Thus we need
$1/2^n<1/1000.$
For positive numbers, a larger denominator gives a smaller unit fraction, so we need $2^n>1000$ .

Now $2^9=512<1000$ while
$2^{10}=1024>1000.$
Stage 9 fails, and every earlier stage has even more white area remaining.
Stage 10 succeeds, so it is the first.
The strict word "more" matters: if a stage left exactly one-thousandth white, it would not qualify.

Answer

Stage 10.

Key idea

In a repeated process, the amount left over may be easier to track than the accumulated amount.
Problem 3: Mark complete
Problem 4 A huge sum with a simple ending

Difficulty: 2 of 3 stars, Challenge

For every positive whole number $n$ , prove that $3^{5^n}+7^{3^{2n}}$ is divisible by 10.

The first exponent is the whole number $5^n$ , and the second exponent is the whole number $3^{2n}$ . You should not attempt to evaluate the enormous powers.
1. Hint 1
  List the last digits of the first four positive powers of 3 and of 7.
2. Hint 2
  Find the remainder of each exponent when divided by 4.
Answer

The sum always ends in 0, so it is divisible by 10.

Full solution

The last digits of successive positive powers of 3 repeat as 3, 9, 7, 1.
Multiplying again by 3 restarts the pattern.
The corresponding cycle for powers of 7 is 7, 9, 3, 1.
In both cases, exponents that leave remainder 1 when divided by 4 give the first last digit in the cycle.

Because 5 leaves remainder 1 when divided by 4, every positive power $5^n$ does too: multiplying two numbers of the form $4k+1$ produces another number of that form.
Thus $3^{5^n}$ ends in 3.

Also
$3^{2n}=(3^2)^n=9^n.$
Since 9 leaves remainder 1 when divided by 4, $9^n$ does as well.
Consequently, $7^{3^{2n}}$ ends in 7.
Adding numbers ending in 3 and 7 produces a number ending in 0.
The argument works for every positive whole number $n$ , independent of how large the powers become.

Answer

The sum always ends in 0, so it is divisible by 10.

Key idea

A repeating cycle can replace a huge calculation with a small remainder.
Problem 4: Mark complete
Problem 5 Roots exactly two apart

Difficulty: 2 of 3 stars, Challenge

Find every positive whole number $n$ for which $\sqrt{n+40}-\sqrt n=2$ .

Your solution must not assume at the start that either square root is a whole number. Explain why your answer is the only possibility.
1. Hint 1
  Let the smaller square root be the side length of a square.
2. Hint 2
  Increasing that side length by 2 increases the area by two strips and one small corner square.
Answer

$n=81$ only.

Full solution

Let the smaller square root be $r=\sqrt n$ .
The larger square root must then be $r+2$ .
Squaring these side lengths gives areas $n$ and $n+40$ , so the area increase is exactly 40.

An increase from a square of side $r$ to one of side $r+2$ adds two strips of area $2r$ each and a corner square of area 4.
The increase is therefore $4r+4$ .
Hence $4r+4=40$ : removing the corner area leaves $4r=36$ , so $r=9$ .
This reasoning applies to real side lengths; it did not assume that $r$ was an integer.

Thus $n=r^2=81$ .
Checking the original condition,
$\sqrt{81+40}-\sqrt{81}=\sqrt{121}-9=11-9=2.$
Every possible solution must give the same strip area and therefore the same side length 9, so there are no other values of $n$ .

Answer

$n=81$ only.

Key idea

A difference of square roots can become a difference of areas.
Problem 5: Mark complete
Problem 6 A square in scientific notation

Difficulty: 2 of 3 stars, Challenge

An integer $k$ is allowed to be positive, zero, or negative. Find every value of $k$ for which $3.6\times10^k$ is a positive whole-number perfect square smaller than 1,000,000.

Give the square root in each case and prove that no other exponent works.
1. Hint 1
  Rewrite the number as $36\times10^{k-1}$ when $k$ is at least 1.
2. Hint 2
  In a perfect square, every prime factor occurs an even number of times.
Answer

$k=1,3,5$ , giving $36=6^2$ , $3600=60^2$ , and $360000=600^2$ .

Full solution

If $k=0$ , the number is 3.6, which is not a whole number.
If $k$ is negative, the number is positive but less than 1, so it also cannot be a positive whole number.
Therefore $k$ must be at least 1.

For such exponents,
$3.6\times10^k=36\times10^{k-1}.$
Its prime factors are $2^{k+1}\times3^2\times5^{k-1}$ .
A whole-number square has an even number of every prime factor, because squaring doubles each count.
Both $k+1$ and $k-1$ are even exactly when $k$ is odd.

The size restriction gives $k\leq5$ : at $k=6$ the value is 3,600,000, and larger exponents only increase it.
The positive odd possibilities are therefore 1, 3, and 5.
They give 36, 3,600, and 360,000, whose roots are 6, 60, and 600 respectively.
These checks establish that each permitted exponent really works.

Answer

$k=1,3,5$ , giving $36=6^2$ , $3600=60^2$ , and $360000=600^2$ .

Key idea

Perfect-square conditions become parity conditions on prime-factor counts.
Problem 6: Mark complete
Problem 7 Three powers make 80

Difficulty: 2 of 3 stars, Challenge

Find every triple of nonnegative integers $a\leq b\leq c$ satisfying $2^a+2^b+2^c=80$ . Repeated exponents are allowed, and $2^0=1$ .

Prove that your list is complete without testing a long list of triples.
1. Hint 1
  The largest term is at least one-third of 80 and is smaller than 80.
2. Hint 2
  Once you know the largest power, bound the larger of the two remaining terms.
Answer

$(a,b,c)=(3,3,6)$ or $(4,5,5)$ .

Full solution

The largest term $2^c$ is at least $80/3$ , because otherwise all three terms together would be less than 80.
It is smaller than 80, because the other terms are positive.
The only powers of 2 in that interval are 32 and 64.

If the largest term is 64, the other two sum to 16.
The larger of those two must be at least 8 and less than 16, so it must be 8.
The remaining term is also 8.
This gives $8+8+64=80$ , with exponents $(3,3,6)$ .

If the largest term is 32, the other two sum to 48.
Their larger term must be at least 24 and at most 32, forcing it to be 32.
The final term is 16.
This gives $16+32+32=80$ , with exponents $(4,5,5)$ .
Both triples satisfy the required order.
The bounds covered every possible largest term, so the list is complete.

Answer

$(a,b,c)=(3,3,6)$ or $(4,5,5)$ .

Key idea

Bounds can reduce an infinite search to a few forced cases.
Problem 7: Mark complete
Problem 8 A square and a cube at once

Difficulty: 3 of 3 stars, Deep challenge

Find every positive whole number $n$ for which $18n$ is a perfect square and $12n$ is a perfect cube.

Describe all answers with a single formula, find the smallest one, and prove both that your formula works and that it includes every answer.
1. Hint 1
  Write $18=2\times3^2$ and $12=2^2\times3$ .
2. Hint 2
  Track separately how many factors of 2, of 3, and of any other prime occur in $n$ .
Answer

All answers are $n=18t^6$ , where $t$ is any positive whole number. The smallest is 18.

Full solution

Suppose $n$ contains $u$ factors of 2 and $v$ factors of 3 in its prime factorization.
For $18n$ to be a square, $u+1$ and $v+2$ must be even.
For $12n$ to be a cube, $u+2$ and $v+1$ must be multiples of 3.

The allowed counts $u$ are therefore 1, 7, 13, ...: they are odd, and adding 2 makes a multiple of 3.
The allowed counts $v$ are 2, 8, 14, ...: they are even, and adding 1 makes a multiple of 3.
Any prime other than 2 or 3 receives no extra factors from 18 or 12, so its count in $n$ must be both even and a multiple of 3, hence a multiple of 6.

Remove one factor of 2 and two factors of 3 from $n$ .
Every remaining prime count is now a multiple of 6, so the remaining number is a sixth power.
Thus $n=18t^6$ for a positive whole number $t$ .

Conversely, this formula gives $18n=324t^6=(18t^3)^2$ and $12n=216t^6=(6t^2)^3$ , so every such number works.
The smallest positive choice is $t=1$ , giving $n=18$ .

Answer

All answers are $n=18t^6$ , where $t$ is any positive whole number. The smallest is 18.

Key idea

Simultaneous square and cube conditions can force a sixth-power structure.
Problem 8: Mark complete
Problem 9 When is the root sum whole?

Difficulty: 3 of 3 stars, Deep challenge

Find every whole number $n$ with $1\leq n\leq129$ such that $\sqrt n+\sqrt{130-n}$ is a whole number.

Do not assume that a whole-number sum automatically means that both square roots are whole numbers. Prove the fact you need, then find every value of $n$ .
1. Hint 1
  Call the whole-number sum $s$ . The difference of the two roots is $(2n-130)/s$ .
2. Hint 2
  First show both roots are rational. Then explain why a rational square root of a whole number must itself be whole.
Answer

$n=9,49,81,121$ . The corresponding sums are 14,16,16,14.

Full solution

Let $r=\sqrt n$ , $t=\sqrt{130-n}$ , and let their sum be the positive whole number $s$ .
Expanding $(r-t)(r+t)$ gives $r^2+rt-tr-t^2$ ; the two middle terms cancel.
Since $(r-t)(r+t)=r^2-t^2=2n-130$ , their difference is $(2n-130)/s$ , a rational number.
Adding and subtracting this difference and the sum $s$ shows that each root is rational.

Now justify the needed root fact.
Write a nonnegative rational root as $p/q$ in lowest terms, with positive whole-number denominator $q$ .
If its square is a whole number, then $p^2$ is a whole-number multiple of $q^2$ .
If $q$ had a prime factor, that prime would also divide $p^2$ and therefore $p$ , contradicting lowest terms.
Thus $q=1$ : the root is whole.

We therefore need two positive whole-number squares adding to 130.
The smaller root is at most 8, since two roots at least 9 would give a square sum at least 162.
For smaller roots 1 through 8, the required other squares are 129, 126, 121, 114, 105, 94, 81, and 66.
Only 121 and 81 are squares, giving the pairs $(3,11)$ and $(7,9)$ .

Allowing either member of each pair to be $\sqrt n$ gives
$n=9,121,49,81.$
Each works, with root sums 14 or 16, and the argument excludes all other values.

Answer

$n=9,49,81,121$ . The corresponding sums are 14,16,16,14.

Key idea

A whole-number result may force hidden rationality; prove that step before enumerating.
Problem 9: Mark complete
Problem 10 A sum of squares that cannot be square

Difficulty: 3 of 3 stars, Deep challenge

For a positive whole number $n$ , let $S=1+4+4^2+\ldots+4^n$ . Each term is a perfect square.

Can $S$ ever be a perfect square? Give a proof that works for every positive whole number $n$ . Checking examples or listing possible last digits alone is not sufficient.
1. Hint 1
  Consider the remainder after division by 8.
2. Hint 2
  First determine the possible remainders of a square: treat an even number and an odd number separately.
Answer

No. For every positive whole number $n$ , $S$ leaves remainder 5 on division by 8, which no perfect square can do.

Full solution

Because $n$ is positive, the sum includes its first two terms, 1 and 4.
Every later term is divisible by 16 and therefore by 8.
Thus $S$ always has the form $8k+5$ for a whole number $k$ : its remainder after division by 8 is 5.

We now prove which remainders a square can have.
If an integer is even, write it as $2m$ .
Its square is $4m^2$ .
When $m$ is even this is divisible by 8; when $m$ is odd it leaves remainder 4.
If an integer is odd, write it as $2m+1$ .
Its square is $4m(m+1)+1$ .
One of the consecutive integers $m,m+1$ is even, so $4m(m+1)$ is divisible by 8.
Every odd square therefore leaves remainder 1.

The only square remainders are 0, 1, and 4, never 5.
This rules out $S$ for every positive $n$ , including $n=1$ , when $S=5$ .
The excluded case $n=0$ would give $S=1$ , which is a square.
The positive condition is essential.

Answer

No. For every positive whole number $n$ , $S$ leaves remainder 5 on division by 8, which no perfect square can do.

Key idea

A fixed remainder can prove impossibility for infinitely many cases at once.
Problem 10: Mark complete