AMC 10 · 2016 · #8
Easy mode Grade 5Multiply 2015 by itself 2016 times, then subtract 2017. The result is a huge number. What is its tens digit — the second digit from the right?
Pick an answer.
AMC 10 2016 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: Find the tens digit of the number $2015^{2016} - 2017$.
Givens: The number is $2015^{2016} - 2017$; $2015^{2016}$ means $2015$ multiplied by itself $2016$ times; Answer choices: (A) $0$, (B) $1$, (C) $3$, (D) $5$, (E) $8$
Unknowns: The tens digit (the second digit from the right) of $2015^{2016} - 2017$
Understand
Restated: Find the tens digit of the number $2015^{2016} - 2017$.
Givens: The number is $2015^{2016} - 2017$; $2015^{2016}$ means $2015$ multiplied by itself $2016$ times; Answer choices: (A) $0$, (B) $1$, (C) $3$, (D) $5$, (E) $8$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #9 Solve an Easier Related Problem, #7 Identify Subproblems
We cannot write out $2015^{2016}$ — it has thousands of digits. But the question only asks for the tens digit, and the tens digit depends only on the last two digits of a number. Tool #9 (Solve an Easier Related Problem) lets us throw away everything except the last two digits at every step, turning a monster power into a two-digit chase. Once the problem is that small, Tool #5 (Look for a Pattern) shows that the last two digits of the powers of $15$ settle into a short repeating cycle, so we never multiply $2016$ times. Tool #7 (Identify Subproblems) splits the job into two clean pieces: first get the last two digits of the power, then subtract $2017$.
Execute — Answer: A
5.NBT.A.1 Step 1 Only the last two digits matter
- The tens digit of a number lives in its second-from-right place, so it is decided entirely by the last two digits.
- That means we can ignore every digit further to the left and just track the last two digits through the whole calculation.
💡 Whatever happens far to the left can never reach down and change the tens place.
4.NBT.B.5 Step 2 Replace 2015 with 15
- When you multiply numbers, the last two digits of the answer depend only on the last two digits of the factors.
- The last two digits of $2015$ are $15$, so $2015^{2016}$ ends in the same two digits as $15^{2016}$.
- We can work with $15$ instead of $2015$.
💡 In multiplication only the tail of each number feeds the tail of the product.
4.OA.C.5 Step 3 Find the pattern in powers of 15
- List the last two digits of the powers of $15$: $15^1$ ends in $15$; $15^2 = 225$ ends in $25$; $15^3 = 15 \times 25 = 375$ ends in $75$; $15^4 = 15 \times 75 = 1125$ ends in $25$.
- From the second power on, the last two digits flip between $25$ and $75$: even powers end in $25$, odd powers end in $75$.
💡 Each extra factor of $15$ just toggles the ending between $25$ and $75$.
4.OA.C.5 Step 4 Apply the pattern at exponent 2016
- The exponent is $2016$, which is even, so by the pattern $15^{2016}$ ends in $25$.
- Therefore $2015^{2016}$ also ends in $25$ — its last two digits are $25$.
💡 An even exponent lands on the $25$ rung of the alternating ladder.
4.NBT.B.4 Step 5 Subtract 2017 and read the tens digit
- Now subtract $2017$.
- Only the last two digits matter, and $2017$ ends in $17$.
- The power ends in $25$, and since the power is enormous there are higher digits to borrow from, so the last two digits of the difference are $25 - 17 = 08$.
- The last two digits are $0$ and $8$, so the tens digit is $0$.
- The answer is (A).
💡 Subtracting a small number only disturbs the tail, and $25$ minus $17$ leaves $08$.
5.NBT.A.1 The tens digit of a number lives in its second-from-right place, so it is decide 4.NBT.B.5 When you multiply numbers, the last two digits of the answer depend only on the 4.OA.C.5 List the last two digits of the powers of $15$: $15^1$ ends in $15$; $15^2 = 225 4.OA.C.5 The exponent is $2016$, which is even, so by the pattern $15^{2016}$ ends in $25 4.NBT.B.4 Now subtract $2017$. Only the last two digits matter, and $2017$ ends in $17$. T Review
Reasonableness: Sanity-check the cycle with a smaller even exponent: $15^4 = 50625$, whose last two digits are $25$, matching the rule for even powers. Subtracting a number ending in $17$ from one ending in $25$ gives an ending of $08$, so a tens digit of $0$ is consistent. The result $0$ is choice (A).
Alternative: Split $2015 = 2000 + 15$ and expand $2015^{2016}$; every term that contains a factor of $2000$ is a multiple of $100$ and cannot affect the last two digits, so only $15^{2016}$ survives mod $100$. Computing $15^{2016} \bmod 100$ gives $25$ as before, and $25 - 17 = 08$, again yielding tens digit $0$.
CCSS standards used (min grade 5)
5.NBT.A.1Recognize that a digit in one place represents ten times as much as to its right (Knowing the tens digit is set only by the last two digits, so we may discard all higher places.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Seeing that only the last two digits of the factors control the last two digits of a product, so $2015$ may be replaced by $15$.)4.OA.C.5Generate a number or shape pattern following a given rule (Finding the repeating cycle in the last two digits of powers of $15$ ($25$ for even powers, $75$ for odd) and applying it to exponent $2016$.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Subtracting $2017$ to get last two digits $25 - 17 = 08$ and reading off the tens digit.)
⭐ If a question only wants the last digits, keep just the last two digits at every step and hunt for the pattern in the powers.
⭐ If a question only wants the last digits, keep just the last two digits at every step and hunt for the pattern in the powers.
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