AMC 10 · 2014 · #25
Grade 8 arithmeticThe number 5867 is between 22013 and 22014. How many pairs of integers (m,n) are there such that 1≤m≤2012 and 5n<2m<2m+2<5n+1?
Pick an answer.
AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Powers of $5$ cut the number line into blocks; the block for index $n$ is the open interval between $5^n$ and $5^{n+1}$. Count the pairs of integers $(m,n)$ with $1\le m\le 2012$ for which $5^n<2^m<2^{m+2}<5^{n+1}$, that is, the blocks that hold three consecutive powers of two $2^m,2^{m+1},2^{m+2}$. You are given that $5^{867}$ lies between $2^{2013}$ and $2^{2014}$.
Givens: $2^{2013}<5^{867}<2^{2014}$; We want integer pairs $(m,n)$ with $1\le m\le 2012$; The condition on a pair is $5^n<2^m<2^{m+2}<5^{n+1}$; Answer choices: (A) $278$, (B) $279$, (C) $280$, (D) $281$, (E) $282$
Unknowns: The number of pairs $(m,n)$ that satisfy the chain of inequalities
Understand
Restated: Powers of $5$ cut the number line into blocks; the block for index $n$ is the open interval between $5^n$ and $5^{n+1}$. Count the pairs of integers $(m,n)$ with $1\le m\le 2012$ for which $5^n<2^m<2^{m+2}<5^{n+1}$, that is, the blocks that hold three consecutive powers of two $2^m,2^{m+1},2^{m+2}$. You are given that $5^{867}$ lies between $2^{2013}$ and $2^{2014}$.
Givens: $2^{2013}<5^{867}<2^{2014}$; We want integer pairs $(m,n)$ with $1\le m\le 2012$; The condition on a pair is $5^n<2^m<2^{m+2}<5^{n+1}$; Answer choices: (A) $278$, (B) $279$, (C) $280$, (D) $281$, (E) $282$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #16 Change Focus / Count the Complement, #7 Identify Subproblems
Chasing individual pairs $(m,n)$ across a range of thousands is hopeless, so the plan is to convert the whole question into a small system of equations. Tool #16 (Change Focus) reframes 'count pairs' as 'count the blocks $(5^n,5^{n+1})$ that hold three powers of two,' because each such block yields exactly one pair. Tool #7 (Identify Subproblems) supplies the two facts that make the count finite: every block holds only $2$ or $3$ powers of two (since $4<5<8$), and the given anchor $2^{2013}<5^{867}<2^{2014}$ fixes how many blocks and how many powers of two are in play. Tool #4 (Introduce a Variable) then names the two block-types $a$ and $b$, writes 'total blocks' and 'total powers of two' as two linear equations, and solves them. The whole difficulty collapses into subtracting one equation from another.
Execute — Answer: B
8.EE.A.1 Step 1 Two or three powers of two per block
- Look at one block, the interval between $5^n$ and $5^{n+1}$.
- Going from its left end to its right end multiplies by $5$.
- One step from a power of two to the next multiplies by $2$.
- Since $2^2=4<5<8=2^3$, a single factor of $5$ is worth more than two doublings but less than three.
- So each block contains either exactly $2$ or exactly $3$ powers of two — never fewer, never more.
💡 Because $5$ sits between $4$ and $8$, one step up in fives equals two or three steps up in twos.
6.EE.B.5 Step 2 Rewrite the target as a 3-power block
- The condition $5^n<2^m<2^{m+2}<5^{n+1}$ packs the three consecutive powers $2^m$, $2^{m+1}$, $2^{m+2}$ strictly inside the block between $5^n$ and $5^{n+1}$.
- So a pair $(m,n)$ exists for a block exactly when that block holds $3$ powers of two, and then $m$ is forced — it is the exponent of the smallest power inside.
- Counting the pairs $(m,n)$ is therefore the very same job as counting the blocks that hold $3$ powers of two.
💡 A block roomy enough for three doublings gives exactly one pair, so count roomy blocks instead of pairs.
6.NS.C.7 Step 3 Count the blocks and the powers of two
- The blocks run from the one between $5^0$ and $5^1$ up to the one between $5^{866}$ and $5^{867}$, one for each $n=0,1,\ldots,866$: that is $867$ blocks.
- The powers of two lying strictly inside this whole span, above $5^0=1$ and below $5^{867}$, are $2^1,2^2,\ldots,2^{2013}$, because the anchor $2^{2013}<5^{867}<2^{2014}$ says $2^{2013}$ is the last power of two below $5^{867}$.
- So $2013$ powers of two are shared out among the $867$ blocks.
💡 Line up the powers of $5$ and $2$ in order, then just read off how many of each fall in the span.
8.EE.C.8 Step 4 Set up two equations and solve
- Let $a$ be the number of blocks holding $3$ powers of two and $b$ the number holding $2$.
- Every block is one kind or the other, so $a+b=867$.
- Adding up the powers of two block by block, the $3$-blocks give $3a$ and the $2$-blocks give $2b$, and together they account for all $2013$ powers, so $3a+2b=2013$.
- Subtracting twice the first equation from the second removes $b$: $(3a+2b)-(2a+2b)=2013-1734$, giving $a=279$.
- Since $a$ counts exactly the valid pairs, the answer is (B) $279$.
💡 Two totals — how many blocks and how many powers — pin down the two unknown counts exactly.
8.EE.A.1 Look at one block, the interval between $5^n$ and $5^{n+1}$. Going from its left 6.EE.B.5 The condition $5^n<2^m<2^{m+2}<5^{n+1}$ packs the three consecutive powers $2^m$ 6.NS.C.7 The blocks run from the one between $5^0$ and $5^1$ up to the one between $5^{86 8.EE.C.8 Let $a$ be the number of blocks holding $3$ powers of two and $b$ the number hol Review
Reasonableness: The solution $a=279$ forces $b=867-279=588$, and both totals check out: $279+588=867$ blocks and $3\cdot 279+2\cdot 588=837+1176=2013$ powers of two. The average block holds $2013/867\approx 2.32$ powers of two, comfortably between $2$ and $3$ and nearer $2$, matching the fact that $2$-blocks ($588$) outnumber $3$-blocks ($279$). Neighbouring choices like $278$ or $280$ would break one of the two exact totals, so $279$ is the only consistent value. Every counted $m$ is at most $2011$, so the bound $m\le 2012$ is respected.
Alternative: Think of it continuously. Each block gains a third power of two at the rate $\log_2 5-2\approx 0.3219$ extra powers per block, since one factor of $5$ equals $\log_2 5\approx 2.3219$ doublings. Over $867$ blocks that is $0.3219\times 867\approx 279$ three-power blocks. This equidistribution view lands on the same $279$ and explains why the answer is close to a third of $867$.
CCSS standards used (min grade 8)
8.EE.A.1Know and apply the properties of integer exponents (Comparing powers of $2$ and $5$ through their exponents to show each block $(5^n,5^{n+1})$ holds either $2$ or $3$ powers of two.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Reading the chain $5^n<2^m<2^{m+2}<5^{n+1}$ as the statement 'this block contains three consecutive powers of two.')6.NS.C.7Understand ordering and absolute value of rational numbers (Ordering the powers of $5$ and $2$ on the number line to count $867$ blocks and, via the anchor $2^{2013}<5^{867}<2^{2014}$, the $2013$ powers of two inside them.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Solving $a+b=867$ and $3a+2b=2013$ to get $a=279$, the number of $3$-power blocks.)
⭐ Each gap between powers of $5$ fits either $2$ or $3$ powers of $2$; two totals — $867$ gaps and $2013$ powers — pin down exactly $279$ of them as the roomy $3$-power gaps.
⭐ Each gap between powers of $5$ fits either $2$ or $3$ powers of $2$; two totals — $867$ gaps and $2013$ powers — pin down exactly $279$ of them as the roomy $3$-power gaps.
More like this
Same archetype — closest grade level first.