Each displayed term is twice the preceding term. Starting with the first term at n=1 gives:

Check: n=1,2,3,4 gives 1,2,4,8. This is the formula under the doubling rule; the finite list alone does not prove that this is the only possible rule.

Write the same sum in reverse and add the two lists term by term. There are n pairs, each totaling n+1.

For four rows, 1+2+3+4=10, agreeing with the formula.

Read each run of identical digits in the previous term. For example, 21 is “one 2, one 1,” giving 1211.

The six-digit block repeats. After 337 complete blocks, the required digit is the fourth digit of 142857, which is 8.

Zero is neither positive nor negative. The claim is false. The correct statement is that every real square is nonnegative.

Use the area formula with r=5.

This is exact; approximately 78.54 square units.

Use the positive root because a length is positive.

Let the two even integers be 2m and 2n, where m and n are integers.

Since m+n is an integer, the sum is a multiple of two, so it is even.

Try 2: the steps give 16,22,11,8. Try 5: they give 40,46,23,20. Conjecture: the result is four times the original number.

This algebra proves the conjecture for every real starting number x.

Base: for \(n=1\), both sides equal one. Assume the sum through \(2k-1\) equals \(k^2\). The next odd number is \(2(k+1)-1=2k+1\), so the sum through it is \(k^2+2k+1=(k+1)^2\). This proves the implication for arbitrary k and completes induction. We assumed only the k case temporarily; we did not assume the desired k+1 conclusion.

Understand: the street map is a 4-by-3 grid. Every direct route has seven steps: four right (R) and three down (D). Plan: choose which three positions contain D.

There are 35 direct routes. Check by adding incoming route counts at each junction: the bottom row becomes 1,4,10,20,35.

There are 3 routes from A to the left end of that segment: one right and two down. After walking right along the segment, 3 routes remain to B: two right and one down.

The café lies along an edge of the grid, not at an intersection; require that edge in the count.

A route contains four steps in one direction and five in the other.

For positive integer m, test the increasing counts: m=1 gives 4, m=2 gives 10, m=3 gives 20, and m=4 gives 35. Thus m=4; later counts are larger.

Undo the last subtraction before undoing the last multiplication. Work backward one day at a time.

John started with ₱290. Check forward:

The last denominator is 999×1000. Rewrite every term so that adjacent terms cancel.

Set a₁=√2 and aₙ₊₁=√(2+aₙ). The first term is less than 2. If aₙ<2, then aₙ₊₁<√4=2, so the sequence stays bounded above by 2. It is increasing: the first step increases and the square-root map preserves inequalities. Thus a limit L exists.

Because the limit is nonnegative, reject −1. The value is 2.

For n=1,2,3,4,5 the maximum counts are 1,2,4,8,16. Each new chord adds one more region than its number of interior crossings. There are C(n,2) chords. Every choice of four points determines exactly one interior crossing, so there are C(n,4) crossings.

The next value is 31, not 32. The first five cases suggest doubling but do not prove it. The no-three-concurrent condition ensures crossings are counted separately.

List every unordered positive-integer triple with product 36 and calculate its sum.

The census taker knows the sum but still cannot decide. Only 13 occurs twice, leaving (1,6,6) and (2,2,9). The statement “my oldest child” supplies a unique oldest child. This eliminates the two six-year-olds, so the ages are 2,2,9. Merely giving the product and saying “there is a unique oldest” would not supply all these clues.

With \(15\) cards, three of each label avoid four matching cards. With \(16\), assuming at most three of each would allow at most \(5\cdot3=15\), a contradiction. Thus \(16\) is both sufficient and minimal. A likely outcome and a guaranteed outcome are different claims.

Count both upright columns, then the bottom circles strictly between them. The corners must not be counted twice.

The first four counts are 2,5,8,11, matching the diagrams.

Caryl counts two columns of n circles and the n−1 bottom interior circles. Darius starts with a full n-by-(n+1) rectangle of circles, then removes the (n−1)-by-(n−1) interior block above the bottom row.

Both decompositions count the same circles.

Let r be the number of red cards in the first group. Because it contains 26 cards, its black count is 26−r. The total black count is 26, so the second group contains 26−(26−r)=r black cards.

If r>0, the ratio is 1:1. If r=0, both counts are zero and the ratio is undefined; the equality of the counts always holds. Random cutting does not change the argument.

One answer is the North Pole. Walking south reaches a parallel one mile from the pole. Walking west changes longitude; walking north one mile then returns to the North Pole.

There are also infinitely many starting points near the South Pole. Choose a parallel whose circumference is 1/k mile, for a positive integer k. Start one mile north of that parallel. The southward leg reaches it; walking one mile west makes exactly k complete circuits; the northward leg retraces the first leg to the start. Thus the North Pole is not the only solution.

• Dave must play the harmonica: all four other choices are excluded.
• Jim cannot use Dave’s instrument, so he plays the saxophone.
• Carol cannot use either of those instruments, leaving the clarinet.
• Sue can choose only flute or trombone; the clue excludes flute, leaving trombone.

The flute is unused. Check that every clue holds and no instrument is assigned twice.

1. Count chords and crossings: \(1+\binom62+\binom64=1+15+15=31\). The initial doubling pattern is not a proof.
2. The compared counts are both zero. Their equality still holds, but \(0/0\) is undefined. State the \(1:1\) ratio only when the common count is positive.

1. No. For \(x=-1\), \(\sqrt{x^2}=1\ne-1\). The correct identity for all real \(x\) is \(\sqrt{x^2}=|x|\).
2. No. At \(x=0\), \(|x|=0\). The universally true statement is \(|x|\ge0\).
3. No. At \(x=-1\), the claim would be \(-2\ge-1\), which is false. Subtracting \(x\) from both sides shows the original inequality holds exactly when \(x\ge0\).
4. No. At \(x=-3\), the left side is \(0\) and the right side is \(6\). Absolute value does not generally distribute over addition.
5. Apply \(A=\pi r^2\). Substituting the given radius gives \(A=\pi\cdot5^2=25\pi\) square units.
6. By the Pythagorean theorem, \(c^2=15^2+8^2=225+64=289\). Length is positive, so \(c=\sqrt{289}=17\) units.

Every move changes the value by an even integer, so all reachable numbers are even. Thus \(15\) is impossible. Parity alone does not prove every even target is reachable, but \(14\) has an explicit construction: \(0\to6\to12\to18\to14\). A necessary condition can rule something out; a construction proves a particular target possible.

Take \(a_n=2^{n-1}\) and \(b_n=2^{n-1}+(n-1)(n-2)(n-3)(n-4)\). At each of \(n=1,2,3,4\), one factor in the added product is zero. At \(n=5\), however, \(a_5=16\) and \(b_5=40\). An explicit recurrence such as \(a_1=1\), \(a_{n+1}=2a_n\) would determine the continuation. The finite observations alone do not establish that recurrence.